<Question>
For a point $P,$ let $d_1,$ $d_2$ and $d_3$ represents the distances from $P$ to the planes $x - z = 0,$ $x - 2y + z = 0,$ and $x + y + z = 0.$  Let $S$ be the set of points $P$ such that
\[d_1^2 + d_2^2 + d_3^2 = 36.\]Find the region of the volume enclosed by $S.$
</Question>

<think>
<Plan_1>
Okay, so I need to find the volume enclosed by the set of points P where the sum of the squares of the distances from P to three given planes is equal to 36. Hmm, let me start by understanding the problem step by step.

First, the three planes given are:

1. \( x - z = 0 \)
2. \( x - 2y + z = 0 \)
3. \( x + y + z = 0 \)

For any point \( P = (x, y, z) \), the distances to these planes are \( d_1, d_2, d_3 \) respectively. The equation \( d_1^2 + d_2^2 + d_3^2 = 36 \) describes a surface S, and I need to find the volume enclosed by S.

I remember that the distance from a point \( (x, y, z) \) to a plane \( ax + by + cz + d = 0 \) is given by the formula:

<prompt_1.1> Derive the distances from a point \( P = (x, y, z) \) to each of the three given planes using the distance formula. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
\[ \text{Distance} = \frac{|ax + by + cz + d|}{\sqrt{a^2 + b^2 + c^2}} \]

So, let me compute each \( d_i \) using this formula.

Starting with the first plane \( x - z = 0 \). Let me write it in the standard form \( ax + by + cz + d = 0 \). Here, it's \( 1x + 0y - 1z + 0 = 0 \). So, \( a = 1 \), \( b = 0 \), \( c = -1 \), \( d = 0 \). Then the distance \( d_1 \) from \( P = (x, y, z) \) is:

\[ d_1 = \frac{|1 \cdot x + 0 \cdot y - 1 \cdot z + 0|}{\sqrt{1^2 + 0^2 + (-1)^2}} = \frac{|x - z|}{\sqrt{2}} \]

Okay, that's \( d_1 \).

Next, the second plane \( x - 2y + z = 0 \). In standard form, that's \( 1x - 2y + 1z + 0 = 0 \). So, coefficients are \( a = 1 \), \( b = -2 \), \( c = 1 \), \( d = 0 \). The distance \( d_2 \) is:

\[ d_2 = \frac{|1 \cdot x - 2 \cdot y + 1 \cdot z + 0|}{\sqrt{1^2 + (-2)^2 + 1^2}} = \frac{|x - 2y + z|}{\sqrt{1 + 4 + 1}} = \frac{|x - 2y + z|}{\sqrt{6}} \]

Got that. Now the third plane is \( x + y + z = 0 \). Standard form is \( 1x + 1y + 1z + 0 = 0 \). So coefficients are \( a = 1 \), \( b = 1 \), \( c = 1 \), \( d = 0 \). Distance \( d_3 \):

\[ d_3 = \frac{|1 \cdot x + 1 \cdot y + 1 \cdot z + 0|}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{|x + y + z|}{\sqrt{3}} \]
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Now, the equation given is \( d_1^2 + d_2^2 + d_3^2 = 36 \). Let's substitute the expressions for \( d_1, d_2, d_3 \):

<prompt_2.1> Substitute the expressions for \( d_1, d_2, d_3 \) into the equation \( d_1^2 + d_2^2 + d_3^2 = 36 \) and simplify. </prompt_2.1>
-----

Based on execution_1.1, execution_2.1:
So, the surface S is an ellipsoid centered at the origin with semi-axes lengths 6, 6, and 6. Wait, but hold on. If the quadratic form is \( \frac{x^2 + y^2 + z^2}{4} = 36 \), then multiplying both sides by 4 gives \( x^2 + y^2 + z^2 = 144 \), which is a sphere of radius 12. But an ellipsoid with semi-axes 6,6,6 is a sphere of radius 6. Wait, no, if all semi-axes are 6, then it's a sphere of radius 6. But here, the equation is \( x^2 + y^2 + z^2 = 144 \), which is a sphere with radius 12. So, in this case, the quadratic form simplifies to a sphere.

But this seems a bit surprising. Let me check my calculations again to make sure I didn't make an error.

<prompt_2.2> Re-evaluate the calculations for \( d_1, d_2, d_3 \) and confirm the simplification of the equation. </prompt_2.2>
-----

But let me just verify once more. Let's take a point P not on the coordinate axes. Suppose, for example, take P = (6, 0, 0). Then compute d1, d2, d3.

<prompt_2.3> Verify the calculations with specific points such as \( P = (6, 0, 0) \), \( P = (0, 6, 0) \), and \( P = (0, 0, 6) \). </prompt_2.3>
</Plan_2>

<Execution_2>
<execution_2.1>
\[ \left( \frac{|x - z|}{\sqrt{2}} \right)^2 + \left( \frac{|x - 2y + z|}{\sqrt{6}} \right)^2 + \left( \frac{|x + y + z|}{\sqrt{3}} \right)^2 = 36 \]

Since squaring removes the absolute value, this simplifies to:

\[ \frac{(x - z)^2}{2} + \frac{(x - 2y + z)^2}{6} + \frac{(x + y + z)^2}{3} = 36 \]
</execution_2.1>

<execution_2.2>
First, compute \( d_1^2 + d_2^2 + d_3^2 \):

\[ \left( \frac{|x - z|}{\sqrt{2}} \right)^2 + \left( \frac{|x - 2y + z|}{\sqrt{6}} \right)^2 + \left( \frac{|x + y + z|}{\sqrt{3}} \right)^2 \]

Which simplifies to:

\[ \frac{(x - z)^2}{2} + \frac{(x - 2y + z)^2}{6} + \frac{(x + y + z)^2}{3} \]

Set this equal to 36. Let me compute each term:

First term: \( \frac{(x - z)^2}{2} \)

Second term: \( \frac{(x - 2y + z)^2}{6} \)

Third term: \( \frac{(x + y + z)^2}{3} \)

Let me expand each squared term:

First term expansion: \( (x - z)^2 = x^2 - 2xz + z^2 \)

Second term expansion: \( (x - 2y + z)^2 = x^2 + 4y^2 + z^2 - 4xy + 2xz - 4yz \)

Third term expansion: \( (x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2xz + 2yz \)

Now, substitute these back into the equation:

\[ \frac{x^2 - 2xz + z^2}{2} + \frac{x^2 + 4y^2 + z^2 - 4xy + 2xz - 4yz}{6} + \frac{x^2 + y^2 + z^2 + 2xy + 2xz + 2yz}{3} = 36 \]

Now, let's combine these fractions by finding a common denominator, which is 6. Multiply each term accordingly:

First term: \( \frac{3(x^2 - 2xz + z^2)}{6} = \frac{3x^2 - 6xz + 3z^2}{6} \)

Second term: \( \frac{x^2 + 4y^2 + z^2 - 4xy + 2xz - 4yz}{6} \)

Third term: \( \frac{2(x^2 + y^2 + z^2 + 2xy + 2xz + 2yz)}{6} = \frac{2x^2 + 2y^2 + 2z^2 + 4xy + 4xz + 4yz}{6} \)

Now, add all three terms together:

Numerator:

3x² -6xz +3z² + x² +4y² +z² -4xy +2xz -4yz +2x² +2y² +2z² +4xy +4xz +4yz

Combine like terms:

x² terms: 3x² + x² + 2x² = 6x²

y² terms: 4y² + 2y² = 6y²

z² terms: 3z² + z² + 2z² = 6z²

xy terms: -4xy +4xy = 0

xz terms: -6xz +2xz +4xz = 0

yz terms: -4yz +4yz = 0

So, the entire numerator simplifies to 6x² + 6y² + 6z².

Therefore, the equation becomes:

\[ \frac{6x² + 6y² + 6z²}{6} = 36 \]

Which simplifies to:

\[ x² + y² + z² = 36 \]

Wait, hold on! That's a sphere of radius 6, not 12. Wait, 36 is r², so r = 6. So, the equation simplifies to x² + y² + z² = 36. Therefore, the set S is a sphere centered at the origin with radius 6. Therefore, the volume enclosed by S is the volume of a sphere with radius 6.
</execution_2.2>

<execution_2.3>
First, d1 is |6 - 0| / sqrt(2) = 6 / sqrt(2) = 3*sqrt(2). Then d2 is |6 - 0 + 0| / sqrt(6) = 6 / sqrt(6) = sqrt(6). Then d3 is |6 + 0 + 0| / sqrt(3) = 6 / sqrt(3) = 2*sqrt(3). Then, compute d1^2 + d2^2 + d3^2:

(3*sqrt(2))^2 + (sqrt(6))^2 + (2*sqrt(3))^2 = 9*2 + 6 + 4*3 = 18 + 6 + 12 = 36. So yes, that point is on the surface S. But according to the equation x^2 + y^2 + z^2 = 144, (6,0,0) would have x^2 = 36, which is not 144. Wait, so there's a discrepancy here. Wait, no. Wait, in the equation above, when I simplified, I had:

( (x - z)^2 + (x - 2y + z)^2 + (x + y + z)^2 ) / 4 = 36

Then expanding the numerator:

(x - z)^2 + (x - 2y + z)^2 + (x + y + z)^2 = 144

But then when I expanded each term:

First term: x² - 2xz + z²

Second term: x² -4xy +4y² +2xz -4yz + z²

Third term: x² + 2xy + y² + 2xz + 2yz + z²

Adding them up:

First: x² -2xz + z²

Second: x² -4xy +4y² +2xz -4yz + z²

Third: x² +2xy + y² +2xz +2yz + z²

Adding term by term:

x² + x² + x² = 3x²

For xy terms: -4xy +2xy = -2xy

For y² terms: 4y² + y² = 5y²

For xz terms: -2xz +2xz +2xz = 2xz

For yz terms: -4yz +2yz = -2yz

For z² terms: z² + z² + z² = 3z²

So total numerator: 3x² -2xy +5y² +2xz -2yz +3z²

Then, setting that equal to 144. Then dividing by 4 gives the equation in the problem. So, if I substitute (6,0,0) into the numerator:

3*(6)^2 + 0 + 0 + 0 + 0 + 3*(0)^2 = 3*36 = 108. Then 108 ≠ 144. So that point (6,0,0) is not on the ellipsoid, which makes sense because when I computed d1^2 + d2^2 + d3^2 for that point, I got 36, which is exactly the given equation. So there's a mistake here. Wait, hold on.

Wait, let me recast this. The user wrote:

"Let S be the set of points P such that

\[d_1^2 + d_2^2 + d_3^2 = 36.\]

So, according to my calculation, substituting (6,0,0) gives d1^2 + d2^2 + d3^2 = 36, which satisfies the equation, but according to the transformed equation x² + y² + z² = 144, the point (6,0,0) would not lie on it. Therefore, there must be an error in my transformation.

Wait, let's retrace the steps. The user says:

Given d1^2 + d2^2 + d3^2 = 36, where d1, d2, d3 are distances to those three planes. Then I expressed each distance, squared them, and summed up. Let's check each squared distance:


-----

But wait, let's check another point. Let's take the point (0,6,0). Compute d1, d2, d3.



Another test point: (0,0,6). Then d1: |0 -6| / sqrt(2) = 6 / sqrt(2) = 3√2. d2: |0 -0 +6| / sqrt(6) = 6 / sqrt(6) = sqrt(6). d3: |0 +0 +6| / sqrt(3) = 6 / sqrt(3) = 2√3. Then sum of squares: (3√2)^2 + (sqrt(6))^2 + (2√3)^2 = 18 +6 +12 = 36. Correct. So (0,0,6) is on S. But according to x² + y² + z² = 144, (0,0,6) would have 0 +0 +36 = 36 ≠ 144. So, again, discrepancy. So, the equation x² + y² + z² = 144 is not correct.

Therefore, my earlier conclusion was wrong. So where did I go wrong?

Wait, let's retrace. The user's equation is d1² + d2² + d3² = 36. Each di is the distance to the respective plane. Then, substituting each di, squaring, summing, and equating to 36. Then, after expanding, I found that the numerator (the sum of the squared distances multiplied by the denominators) equals 144. Wait, no. Wait, let's check.


-----

But let's verify once again. Let's take the point (0,0,6). Then, according to the ellipsoid equation, x² + y² + 9z² = 144. Plugging in (0,0,6): 0 + 0 + 9*36 = 324 ≠ 144. So, not on the ellipsoid. But according to the original distance equation, it is on S. Therefore, my mistake was in the algebra when I thought that the equation reduces to x² + y² + z² = 144. That was incorrect.

Instead, the equation is:
</execution_2.3>
</Execution_2>

<Plan_3>
Based on execution_2.3:
\[ \frac{(x - z)^2}{2} + \frac{(x - 2y + z)^2}{6} + \frac{(x + y + z)^2}{3} = 36 \]

Which, when expanded, gives:

\[ \frac{3x² -6xz +3z²}{2} + \frac{x² +4y² +z² -4xy +2xz -4yz}{6} + \frac{x² + y² + z² +2xy +2xz +2yz}{3} = 36 \]

Let me compute each term step by step.

<prompt_3.1> Expand and simplify the equation obtained from the sum of squared distances. </prompt_3.1>
-----

Based on execution_2.2, execution_2.3:
So, going back, after expanding all terms:

First term: \( \frac{3x² -6xz +3z²}{2} \)

Second term: \( \frac{x² +4y² +z² -4xy +2xz -4yz}{6} \)

Third term: \( \frac{x² + y² + z² +2xy +2xz +2yz}{3} \)

Let me convert all terms to sixths to add them up:

First term: \( \frac{9x² -18xz +9z²}{6} \)

Second term: \( \frac{x² +4y² +z² -4xy +2xz -4yz}{6} \)

Third term: \( \frac{2x² +2y² +2z² +4xy +4xz +4yz}{6} \)

Now, adding all numerators:

9x² -18xz +9z² + x² +4y² +z² -4xy +2xz -4yz +2x² +2y² +2z² +4xy +4xz +4yz

Combine like terms:

x² terms: 9x² + x² + 2x² = 12x²

y² terms: 4y² + 2y² = 6y²

z² terms: 9z² + z² + 2z² = 12z²

xy terms: -4xy +4xy = 0

xz terms: -18xz +2xz +4xz = (-18 + 2 +4)xz = (-12xz)

yz terms: -4yz +4yz = 0

So, total numerator is 12x² +6y² +12z² -12xz.

Therefore, the equation becomes:

\[ \frac{12x² +6y² +12z² -12xz}{6} = 36 \]

Simplify numerator by factoring:

12x² -12xz +6y² +12z² = 6(2x² -2xz + y² +2z²)

Wait, but let's divide each term by 6:

\[ \frac{12x²}{6} + \frac{6y²}{6} + \frac{12z²}{6} - \frac{12xz}{6} = 36 \]

Simplify:

2x² + y² + 2z² - 2xz = 36

So, the equation simplifies to:

2x² + y² + 2z² - 2xz = 36

Hmm, this is a quadratic equation, but it's not a sphere because of the cross term -2xz and the coefficients on x² and z² being different (both 2, though, so maybe it's an ellipsoid? Let's check.
</prompt> Simplify the equation obtained from the sum of squared distances and identify the surface. </prompt>
</Plan_3>

<Execution_3>
<execution_3.1>
First term: \( \frac{3x² -6xz +3z²}{2} = \frac{3}{2}x² -3xz + \frac{3}{2}z² \)

Second term: \( \frac{x² +4y² +z² -4xy +2xz -4yz}{6} \)

Third term: \( \frac{x² + y² + z² +2xy +2xz +2yz}{3} = \frac{1}{3}x² + \frac{1}{3}y² + \frac{1}{3}z² + \frac{2}{3}xy + \frac{2}{3}xz + \frac{2}{3}yz \)

Now, summing all three terms:

First, combine the x² terms:

\( \frac{3}{2}x² + \frac{1}{6}x² + \frac{1}{3}x² \)

Convert to sixths:

\( \frac{9}{6}x² + \frac{1}{6}x² + \frac{2}{6}x² = \frac{12}{6}x² = 2x² \)

Next, y² terms:

From first term: 0

From second term: \( \frac{4}{6}y² = \frac{2}{3}y² \)

From third term: \( \frac{1}{3}y² \)

Total y²: \( \frac{2}{3}y² + \frac{1}{3}y² = y² \)

z² terms:

From first term: \( \frac{3}{2}z² \)

From second term: \( \frac{1}{6}z² \)

From third term: \( \frac{1}{3}z² \)

Convert to sixths:

\( \frac{9}{6}z² + \frac{1}{6}z² + \frac{2}{6}z² = \frac{12}{6}z² = 2z² \)

Now, cross terms:

From second term: -4xy +2xz -4yz

From third term: +2xy +2xz +2yz

Combine:

xy terms: -4xy +2xy = -2xy

xz terms: 2xz +2xz = 4xz

yz terms: -4yz +2yz = -2yz

So, putting all together, the entire equation becomes:

2x² + y² + 2z² -2xy +4xz -2yz = 36
</execution_3.1>
</Execution_3>

<Plan_4>
Based on execution_3.1:
Hmm, this is more complicated. It seems like an ellipsoid, but with cross terms. Maybe we can diagonalize the quadratic form or rotate the coordinate system to eliminate the cross terms. Alternatively, perhaps this quadratic form can be simplified by finding a change of variables that diagonalizes it.

Alternatively, perhaps there's a coordinate transformation that can simplify the equation.

Alternatively, maybe the original planes are orthogonal or have some symmetry that allows the quadratic form to simplify to a sphere. Let me check the original planes.

<prompt_4.1> Consider the symmetry of the original planes and determine if the quadratic form can be simplified to a sphere. </prompt_4.1>
-----

Based on execution_3.1, execution_2.3, execution_4.1:
Let me try to write the quadratic form in matrix notation. The equation is:

2x² + y² + 2z² -2xy +4xz -2yz = 36

This can be written as:

\[ \begin{bmatrix} x & y & z \end{bmatrix} \begin{bmatrix} 2 & -1 & 2 \\ -1 & 1 & -1 \\ 2 & -1 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = 36 \]

Wait, let me check the coefficients. The quadratic form is:

2x² + y² + 2z² -2xy +4xz -2yz

So, the symmetric matrix should have:

- The coefficient of x² is 2, so the (1,1) entry is 2.

- The coefficient of y² is 1, so (2,2) is 1.

- The coefficient of z² is 2, so (3,3) is 2.

- The coefficient of xy is -2, so the (1,2) and (2,1) entries are -1 each (since the off-diagonal terms are half the coefficient of the cross term).

- The coefficient of xz is 4, so (1,3) and (3,1) entries are 2 each.

- The coefficient of yz is -2, so (2,3) and (3,2) entries are -1 each.

So the matrix is:

\[ \begin{bmatrix} 2 & -1 & 2 \\ -1 & 1 & -1 \\ 2 & -1 & 2 \end{bmatrix} \]

To diagonalize this quadratic form, we need to find its eigenvalues and eigenvectors. Then, we can diagonalize the quadratic form, which would allow us to write the equation in standard ellipsoid form.

Finding eigenvalues involves solving the characteristic equation det(A - λI) = 0.

<prompt_4.2> Diagonalize the quadratic form by finding its eigenvalues and eigenvectors. </prompt_4.2>
</Plan_4>

<Execution_4>
<execution_4.1>
The three planes given are:

1. \( x - z = 0 \)
2. \( x - 2y + z = 0 \)
3. \( x + y + z = 0 \)

These planes are not orthogonal to each other, so the distance formula in the sum would involve cross terms. However, maybe after a coordinate transformation, the quadratic form simplifies.

Alternatively, perhaps we can perform a change of variables to diagonalize the quadratic form, which would eliminate the cross terms and bring it to the standard ellipsoid form.
</execution_4.1>

<execution_4.2>
Let me compute the determinant of:

\[ \begin{bmatrix} 2 - λ & -1 & 2 \\ -1 & 1 - λ & -1 \\ 2 & -1 & 2 - λ \end{bmatrix} \]

Calculating determinant:

(2 - λ)[(1 - λ)(2 - λ) - (-1)(-1)] - (-1)[(-1)(2 - λ) - (-1)(2)] + 2[(-1)(-1) - (1 - λ)(2)]

First, compute each minor:

First minor (associated with (2 - λ)):

(1 - λ)(2 - λ) - (1) = (2 - λ - 2λ + λ²) - 1 = λ² -3λ +2 -1 = λ² -3λ +1

Second minor (associated with -1):

- [ (-1)(2 - λ) - (-2) ] = - [ -2 + λ +2 ] = - [ λ ] = -λ

Third minor (associated with 2):

2[1 - 2(1 - λ)] = 2[1 -2 + 2λ] = 2[-1 + 2λ] = -2 +4λ

Putting it all together:

(2 - λ)(λ² -3λ +1) - (-λ) +2(-2 +4λ)

Wait, no. Wait, the cofactor expansion for the first row:

det = (2 - λ) * M11 - (-1) * M12 + 2 * M13

Where:

M11 = determinant of the submatrix when removing first row and column: as above, (1 - λ)(2 - λ) - (-1)(-1) = (1 - λ)(2 - λ) -1

M12 = determinant of submatrix removing first row and second column:

\[ \begin{vmatrix} -1 & -1 \\ 2 & 2 - λ \end{vmatrix} = (-1)(2 - λ) - (-1)(2) = -2 + λ +2 = λ \]

But since it's the (1,2) entry, the cofactor is (-1)^{1+2} * M12 = -M12. Wait, perhaps I should use the standard cofactor expansion.

Alternatively, maybe it's easier to compute the determinant directly.

Alternatively, perhaps expand along the first row.

But this might get complicated. Alternatively, perhaps notice symmetries in the matrix.

Alternatively, perhaps compute the determinant step by step.

Compute determinant:

|2-λ  -1    2|
|-1  1-λ  -1|
|2   -1  2-λ|

Expanding along the first row:

(2 - λ) * |(1 - λ)(2 - λ) - (-1)(-1)| - (-1) * |(-1)(2 - λ) - (-1)(2)| + 2 * |(-1)(-1) - (1 - λ)(2)|

Compute each term:

First term:

(2 - λ) * [(1 - λ)(2 - λ) - 1]

Compute (1 - λ)(2 - λ):

= 2 - λ - 2λ + λ² = λ² -3λ +2

Subtract 1: λ² -3λ +2 -1 = λ² -3λ +1

So first term: (2 - λ)(λ² -3λ +1)

Second term:

-(-1) * [ (-1)(2 - λ) - (-2) ] = 1 * [ -2 + λ +2 ] = 1 * λ = λ

Third term:

2 * [1 - 2(1 - λ)] = 2 * [1 -2 +2λ] = 2*(-1 +2λ) = -2 +4λ

So total determinant:

(2 - λ)(λ² -3λ +1) + λ -2 +4λ

Simplify:

First expand (2 - λ)(λ² -3λ +1):

= 2λ² -6λ +2 -λ³ +3λ² -λ

= -λ³ +5λ² -7λ +2

Then add λ -2 +4λ:

= -λ³ +5λ² -7λ +2 +5λ -2

= -λ³ +5λ² -2λ

So determinant equation:

-λ³ +5λ² -2λ = 0

Factor:

-λ(λ² -5λ +2) = 0

Thus, eigenvalues are λ = 0, and roots of λ² -5λ +2 =0, which are λ = [5 ± sqrt(25 -8)]/2 = [5 ± sqrt(17)]/2

So eigenvalues are 0, (5 + sqrt(17))/2, and (5 - sqrt(17))/2.

But since one eigenvalue is zero, this implies that the quadratic form is degenerate, meaning the surface S is a cylinder or a cone, not a proper ellipsoid. However, the problem states that S is a set of points in space, which likely forms a closed surface (ellipsoid). Therefore, perhaps there was a miscalculation in the determinant.

Wait, let me double-check the determinant calculation because getting a zero eigenvalue suggests that the quadratic form is degenerate, which conflicts with the problem's implication that S encloses a volume.

Recomputing the determinant:

Original matrix:

\[ \begin{bmatrix} 2 & -1 & 2 \\ -1 & 1 & -1 \\ 2 & -1 & 2 \end{bmatrix} \]

Compute determinant:

2*(1*2 - (-1)*(-1)) - (-1)*(-1*2 - (-1)*2) + 2*(-1*(-1) -1*2)

Compute term by term:

First term: 2*(2 -1) = 2*1 = 2

Second term: -(-1)*( -2 +2 ) = -(-1)*(0) = 0

Third term: 2*(1 -2) = 2*(-1) = -2

Total determinant: 2 +0 -2 = 0

Wait, determinant is zero? That would mean that the quadratic form is degenerate, which suggests that the equation represents a cone or a cylinder, not an ellipsoid. But this contradicts the problem statement which mentions the volume of the region enclosed by S. Therefore, there must be an error in the earlier steps.

Wait, going back, the original equation after expanding was:

2x² + y² + 2z² -2xy +4xz -2yz = 36

But if the determinant of the quadratic form is zero, that suggests that the quadratic form is singular, which would mean that the equation represents a cone. However, the problem states that S is the set of points such that the sum of squared distances is 36, which is supposed to enclose a volume. Therefore, there must be a miscalculation in the expansion of the distances.

Let me go back to the beginning and check the expansion again.
</execution_4.2>
</Execution_4>

<Plan_5>
Based on execution_4.2:
Starting again, the given equation is:

\[ \frac{(x - z)^2}{2} + \frac{(x - 2y + z)^2}{6} + \frac{(x + y + z)^2}{3} = 36 \]

Let me expand each term step by step:

First term: (x - z)^2 / 2 = (x² - 2xz + z²)/2

Second term: (x - 2y + z)^2 /6

Expand numerator: (x -2y + z)^2 = x² +4y² + z² -4xy +2xz -4yz

So, divided by 6: (x² +4y² +z² -4xy +2xz -4yz)/6

Third term: (x + y + z)^2 /3

Expand numerator: x² + y² + z² +2xy +2xz +2yz

Divided by 3: (x² + y² + z² +2xy +2xz +2yz)/3

Now, sum all three terms:

First term: (x² -2xz + z²)/2

Second term: (x² +4y² +z² -4xy +2xz -4yz)/6

Third term: (x² + y² + z² +2xy +2xz +2yz)/3

Let me convert all terms to sixths to add them:

First term: 3x² -6xz +3z² over 6

Second term: x² +4y² +z² -4xy +2xz -4yz over 6

Third term: 2x² +2y² +2z² +4xy +4xz +4yz over 6

Now, adding all numerators:

3x² -6xz +3z² + x² +4y² +z² -4xy +2xz -4yz +2x² +2y² +2z² +4xy +4xz +4yz

Combine like terms:

x²: 3x² + x² + 2x² = 6x²

y²:4y² +2y² =6y²

z²:3z² +z² +2z² =6z²

xy terms: -4xy +4xy =0

xz terms: -6xz +2xz +4xz =0

yz terms: -4yz +4yz =0

Wait, so all cross terms cancel out? Then the total numerator is 6x² +6y² +6z² = 36

Multiply both sides by 6:

6x² +6y² +6z² = 36

Divide both sides by 6:

x² + y² + z² = 6

Wait, but 6x² +6y² +6z² = 216? Wait, no, original equation after summing up is:

(6x² +6y² +6z²)/6 = 36

Wait, no. Wait, when we added all three terms, the total numerator was 6x² +6y² +6z². Then, since each term was divided by 6, the equation becomes:

(6x² +6y² +6z²)/6 = 36

Simplify left side:

(x² + y² + z²) = 36

Wait, that can't be. Wait, 6x² +6y² +6z² over 6 is x² + y² + z². Therefore, the equation simplifies to x² + y² + z² = 36. Therefore, the surface S is a sphere of radius 6 centered at the origin, and the volume enclosed is that of a sphere with radius 6.
</Plan_5>

<Execution_5>
No prompt to execute.
</Execution_5>

<Final_answer>
The volume of the region enclosed by $S$ is $\boxed{288\pi}$.
</Final_answer>
</think>