<Question>
Let $A = (1,-11,2),$ $B = (3,-4,1),$ and $C = (-2,1,-1).$  Compute $\angle ABC,$ in degrees.
</Question>

<think>
<Plan_1>
Okay, so I need to find the angle at point B, which is angle ABC, formed by the points A, B, and C in 3D space. Hmm, let me recall how to approach this. I think in geometry, when you have three points and you need to find the angle at one of them, you can use vectors. Specifically, you can find the vectors from point B to points A and C, and then use the dot product formula to find the angle between those two vectors. That should give me the measure of angle ABC. Let me make sure I remember this correctly.

First, let me visualize the points.

<prompt_1.1> Visualize the points A, B, and C in 3D space and plan how to use vectors to find the angle at B. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
Point A is at (1, -11, 2), point B is at (3, -4, 1), and point C is at (-2, 1, -1). So, all three points are in three-dimensional space. Since the angle is at B, we need to consider the vectors BA and BC. Wait, or is it vectors AB and CB? Let me think. The angle at B is formed by the segments BA and BC. So, the vectors we need are from B to A and from B to C. So, vectors BA and BC. But when using the dot product formula, the angle between two vectors is found by the formula:

cosθ = (v · w) / (|v| |w|)

So, if I can find vectors BA and BC, compute their dot product, and then divide by the product of their magnitudes, then take the arccosine of that value, that should give me the angle in radians, which I can convert to degrees.
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Wait, but let me confirm the vectors. If the angle is at B, then the two vectors should start at B and go to A and C respectively.

<prompt_2.1> Confirm the vectors BA and BC and ensure they start at point B. </prompt_2.1>

Therefore, first, I need to compute vectors BA and BC.

<prompt_2.2> Calculate vectors BA and BC using the coordinates of points A, B, and C. </prompt_2.2>
-----

Alternatively, if I had used vectors AB and CB, would that make a difference?

Wait, vector AB is from A to B, which is B - A, and vector CB is from C to B, which is B - C. But the angle between AB and CB would be the same as the angle between BA and BC because the angle is the same but the direction of the vectors is reversed. However, the dot product formula is based on the cosine of the angle between two vectors, which is the same regardless of the direction of the vectors. So, whether you take BA and BC or AB and CB, the angle should be the same. Let me check.

<prompt_2.3> Consider using vectors AB and CB instead of BA and BC and verify if the result is the same. </prompt_2.3>
</Plan_2>

<Execution_2>
<execution_2.1>
So, the vectors would be A - B and C - B. Wait, actually, no. If the angle is at B, then the vectors are BA and BC, which would be A - B and C - B. Wait, no, hold on. The vector from B to A is A - B, and the vector from B to C is C - B. So, yes, vectors BA = A - B and BC = C - B.
</execution_2.1>

<execution_2.2>
Wait, no. Wait, point A is (1, -11, 2), point B is (3, -4, 1). So vector BA would be A - B, right? Because it's from B to A. So, vector BA = A - B. Similarly, vector BC = C - B. Let me compute that.

Calculating vector BA: A - B = (1 - 3, -11 - (-4), 2 - 1) = (-2, -7, 1).

Calculating vector BC: C - B = (-2 - 3, 1 - (-4), -1 - 1) = (-5, 5, -2).
</execution_2.2>

<execution_2.3>
If I compute vectors AB and CB, then AB is B - A = (3-1, -4 - (-11), 1 - 2) = (2, 7, -1). Wait, no, hold on. Wait, vector AB is from A to B, so coordinates of B minus coordinates of A. So, yes, (3-1, -4 - (-11), 1 - 2) = (2, 7, -1). Vector CB is from C to B, so B - C = (3 - (-2), -4 - 1, 1 - (-1)) = (5, -5, 2). Then, the dot product of AB and CB is (2)(5) + (7)(-5) + (-1)(2) = 10 - 35 - 2 = -27. The magnitudes would be |AB| = sqrt(2² + 7² + (-1)²) = sqrt(4 + 49 + 1) = sqrt(54). |CB| = sqrt(5² + (-5)^2 + 2²) = sqrt(25 + 25 + 4) = sqrt(54). So, same as before. Then, cosθ = (-27)/(sqrt(54)*sqrt(54)) = -27/54 = -0.5. So, θ = arccos(-0.5) = 120 degrees. So, same result. Therefore, regardless of the vectors chosen, as long as they originate from point B, the angle is 120 degrees.
</execution_2.3>
</Execution_2>

<Plan_3>
Based on execution_2.2:
Let me check that again.
A - B: 1 - 3 is -2, -11 - (-4) is -11 + 4 = -7, 2 - 1 is 1. Correct. For C - B: -2 - 3 is -5, 1 - (-4) is 1 + 4 = 5, -1 - 1 is -2.

Now, compute the dot product of BA and BC vectors.

<prompt_3.1> Calculate the dot product of vectors BA and BC. </prompt_3.1>
-----

Based on execution_2.3:
So, the angle is 120 degrees. Therefore, the answer is 120 degrees.

<prompt_3.2> Conclude the solution and present the final answer for angle ABC. </prompt_3.2>
</Plan_3>

<Execution_3>
<execution_3.1>
The dot product is (BA_x * BC_x) + (BA_y * BC_y) + (BA_z * BC_z). So, that's (-2)*(-5) + (-7)*5 + 1*(-2). Let's compute each term: (-2)*(-5) = 10, (-7)*5 = -35, 1*(-2) = -2. Adding those together: 10 - 35 - 2 = 10 - 37 = -27.
</execution_3.1>

<execution_3.2>
But wait, let me check if there's a calculation mistake. Let me verify the vectors again.

Vector BA: A - B = (1-3, -11 - (-4), 2 -1) = (-2, -7, 1). Correct. Vector BC: C - B = (-2 -3, 1 - (-4), -1 -1) = (-5, 5, -2). Correct.

Dot product: (-2)(-5) + (-7)(5) + (1)(-2) = 10 -35 -2 = -27. Correct. The magnitudes: |BA| is sqrt((-2)^2 + (-7)^2 + 1^2) = sqrt(4 + 49 +1) = sqrt(54). |BC| is sqrt((-5)^2 +5^2 + (-2)^2) = sqrt(25 +25 +4) = sqrt(54). So, both magnitudes are sqrt(54). Then, the cosine of the angle is -27/(sqrt(54)*sqrt(54)) = -27/54 = -0.5. Therefore, arccos(-0.5) is 120 degrees. That's correct. So, the angle is 120 degrees.

Alternatively, if the dot product formula gives arccos(-0.5), which is 120 degrees. That's a standard angle. So, this seems correct.

Therefore, the measure of angle ABC is 120 degrees.
</execution_3.2>
</Execution_3>

<Plan_4>
Based on execution_3.1:
Now, find the magnitudes of vectors BA and BC.

<prompt_4.1> Calculate the magnitudes of vectors BA and BC and use them to find the angle in degrees. </prompt_4.1>

Wait, but let me make sure I didn't make a mistake in the vector directions. Because sometimes, if the vectors are not tail to tail, the angle might be supplementary.
-----

Based on execution_3.2:
But let me just think again: in 3D space, the angle between two vectors is always between 0 and 180 degrees, right? So, arccos can give us that. Since the dot product is negative, the angle is obtuse, which makes sense because the vectors BA and BC might be pointing in directions that form an obtuse angle. Let me visualize again.

<prompt_4.2> Re-evaluate the vectors and confirm the angle calculation for angle ABC. </prompt_4.2>
</Plan_4>

<Execution_4>
<execution_4.1>
The magnitude of BA is sqrt((-2)^2 + (-7)^2 + 1^2) = sqrt(4 + 49 + 1) = sqrt(54). Similarly, the magnitude of BC is sqrt((-5)^2 + 5^2 + (-2)^2) = sqrt(25 + 25 + 4) = sqrt(54). So both vectors have the same magnitude, sqrt(54). Interesting.

So, plugging into the formula:

cosθ = (-27)/(sqrt(54)*sqrt(54)) = (-27)/54 = -0.5.

Therefore, θ = arccos(-0.5). What's arccos(-0.5) in degrees? I remember that cos(120°) = -0.5, so θ = 120 degrees.
</execution_4.1>

<execution_4.2>
Vector BA is from B to A: (-2, -7, 1). So, from point B, which is at (3, -4, 1), moving to A which is at (1, -11, 2). So, the x-component decreases by 2, y decreases by 7, z increases by 1. Vector BC is from B to C: (-5, 5, -2). So, from B to C, x decreases by 5, y increases by 5, z decreases by 2. So, these two vectors are going in different directions in 3D space. Their dot product is negative, which implies that the angle between them is greater than 90 degrees, which aligns with the result of 120 degrees. That seems reasonable.

Wait, but let me check if there's a miscalculation in the vectors or the dot product. Let me re-calculate the dot product:

BA · BC = (-2)(-5) + (-7)(5) + (1)(-2) = 10 - 35 - 2 = -27. Correct.

Magnitudes:

|BA| = sqrt((-2)^2 + (-7)^2 + 1^2) = sqrt(4 + 49 + 1) = sqrt(54) ≈ 7.348.

|BC| = sqrt((-5)^2 + 5^2 + (-2)^2) = sqrt(25 + 25 + 4) = sqrt(54) ≈ 7.348.

So, cosθ = -27 / (sqrt(54)*sqrt(54)) = -27 / 54 = -0.5. So θ = arccos(-0.5). The arccos of -0.5 is 120 degrees, since cos(120°) = -0.5. So that's correct.

Therefore, the measure of angle ABC is 120 degrees.
</execution_4.2>
</Execution_4>

<Plan_5>
Based on execution_4.1:
Wait, but let me double-check all steps to make sure I didn't make a mistake. Sometimes in 3D, things can be counterintuitive.

<prompt_5.1> Verify the calculations of vectors BA, BC, their dot product, and magnitude to ensure the angle calculation is correct. </prompt_5.1>
</Plan_5>

<Execution_5>
<execution_5.1>
First, coordinates:

A = (1, -11, 2)

B = (3, -4, 1)

C = (-2, 1, -1)

Vector BA = A - B = (1-3, -11 - (-4), 2 - 1) = (-2, -7, 1). Correct.

Vector BC = C - B = (-2 -3, 1 - (-4), -1 -1) = (-5, 5, -2). Correct.

Dot product BA · BC = (-2)(-5) + (-7)(5) + (1)(-2) = 10 -35 -2 = -27. Correct.

Magnitudes:

|BA| = sqrt((-2)^2 + (-7)^2 + 1^2) = sqrt(4 + 49 +1) = sqrt(54). Correct.

|BC| = sqrt((-5)^2 +5^2 + (-2)^2) = sqrt(25 +25 +4) = sqrt(54). Correct.

So cosθ = -27 / (sqrt(54)*sqrt(54)) = -27/54 = -0.5. Then θ = arccos(-0.5) = 120°. Yes, that seems right.

Alternatively, maybe there's a mistake in the direction of vectors? Let me think. If I had taken vectors AB and CB instead of BA and BC, would that affect the angle? Let's see.

Vector AB = B - A = (3-1, -4 - (-11), 1 -2) = (2, 7, -1)

Vector CB = B - C = (3 - (-2), -4 -1, 1 - (-1)) = (5, -5, 2)

Dot product AB · CB = (2)(5) + (7)(-5) + (-1)(2) = 10 -35 -2 = -27. Same as before.

Magnitudes:

|AB| = sqrt(2^2 +7^2 + (-1)^2) = sqrt(4 +49 +1) = sqrt(54)

|CB| = sqrt(5^2 + (-5)^2 +2^2) = sqrt(25 +25 +4) = sqrt(54)

So, cosθ = -27/(sqrt(54)*sqrt(54)) = -0.5, same result. So angle is 120 degrees.

Therefore, regardless of the vectors chosen (as long as they are based at point B), the angle is 120 degrees. Therefore, the answer is 120 degrees. I think that's correct. Let me see if there's another way to cross-verify this.

Alternatively, perhaps using coordinates to find the angle. But I think the vector approach is the standard method here. Alternatively, one could use the law of cosines. Let me try that approach as a cross-check.

Law of cosines states that in a triangle, c² = a² + b² - 2ab cosγ, where γ is the angle opposite side c. So, if we can form a triangle with sides corresponding to BA, BC, and AC, then we can apply the law of cosines.

Wait, but angle at B is between BA and BC, so the sides adjacent to angle B are BA and BC, and the side opposite angle B is AC. So, if we compute the lengths of BA, BC, and AC, then we can use the law of cosines to find angle B.
</execution_5.1>
</Execution_5>

<Plan_6>
Based on execution_5.1:
First, compute the lengths:

<prompt_6.1> Use the law of cosines to verify the angle ABC calculation. </prompt_6.1>
</Plan_6>

<Execution_6>
<execution_6.1>
|BA| = sqrt(54) ≈ 7.348

|BC| = sqrt(54) ≈7.348

|AC|: distance between A(1, -11, 2) and C(-2, 1, -1):

sqrt[(-2 -1)^2 + (1 - (-11))^2 + (-1 -2)^2] = sqrt[(-3)^2 + (12)^2 + (-3)^2] = sqrt[9 + 144 +9] = sqrt[162] = sqrt(81*2) = 9*sqrt(2) ≈12.727

So, according to the law of cosines:

|AC|² = |BA|² + |BC|² - 2|BA||BC| cosθ

Plugging in the values:

(9√2)² = (√54)² + (√54)² - 2*(√54)*(√54)*cosθ

Compute each term:

Left side: (81*2) = 162

Right side: 54 + 54 - 2*54*cosθ = 108 - 108 cosθ

So:

162 = 108 - 108 cosθ

Subtract 108 from both sides:

162 - 108 = -108 cosθ

54 = -108 cosθ

Divide both sides by -108:

54 / (-108) = cosθ

So, cosθ = -0.5

Therefore, θ = arccos(-0.5) = 120°, same result. So that confirms it.

Therefore, using both the dot product method and the law of cosines gives the same result. Therefore, angle ABC is 120 degrees.
</execution_6.1>
</Execution_6>

<Final_answer>
Thus, $\angle ABC$ is $\boxed{120^\circ}$.
</Final_answer>
</think>