Okay, so I need to find a vector v such that when I take the dot product of a and v, I get 2, and when I take the cross product of a and v, I get the vector [1, -2, 1]. The vector a is given as [1, 1, 1]. Hmm, let's see. 

First, I remember that the dot product of two vectors is the sum of the products of their corresponding components. So, if a is [1, 1, 1] and v is [v1, v2, v3], then the dot product a · v is 1*v1 + 1*v2 + 1*v3, which should equal 2. So that gives me the equation v1 + v2 + v3 = 2. That's one equation.

Next, the cross product. The cross product of a and v is given by the determinant of a matrix with the standard basis vectors i, j, k in the first row, the components of a in the second row, and the components of v in the third row. Let me write that out:

a × v = |i j k|
        |1 1 1|
        |v1 v2 v3|

Calculating this determinant, the cross product should be:

i*(1*v3 - 1*v2) - j*(1*v3 - 1*v1) + k*(1*v2 - 1*v1)

Which simplifies to:

(i*(v3 - v2)) - (j*(v3 - v1)) + (k*(v2 - v1))

And this should equal the vector [1, -2, 1]. So, breaking that down component-wise:

First component (i direction): v3 - v2 = 1

Second component (j direction): -(v3 - v1) = -2

Third component (k direction): v2 - v1 = 1

So now I have three equations from the cross product:

1. v3 - v2 = 1

2. -(v3 - v1) = -2 → which simplifies to v3 - v1 = 2

3. v2 - v1 = 1

And from the dot product, we have:

4. v1 + v2 + v3 = 2

So now I have four equations with three variables. Wait, that might be a problem. Let me check if these equations are consistent.

Let me write them out again:

From cross product:

Equation 1: v3 - v2 = 1

Equation 2: v3 - v1 = 2

Equation 3: v2 - v1 = 1

From dot product:

Equation 4: v1 + v2 + v3 = 2

So let's see if we can solve these equations step by step.

First, from Equation 3: v2 = v1 + 1

From Equation 1: v3 = v2 + 1. But since v2 = v1 + 1, then v3 = (v1 + 1) + 1 = v1 + 2

From Equation 2: v3 - v1 = 2. But v3 is v1 + 2, so (v1 + 2) - v1 = 2 → 2 = 2. That's always true, so Equation 2 is redundant given Equations 1 and 3. So we have v3 = v1 + 2 and v2 = v1 + 1.

Now plug these into Equation 4: v1 + (v1 + 1) + (v1 + 2) = 2

Combine like terms: 3v1 + 3 = 2 → 3v1 = -1 → v1 = -1/3

Then, v2 = (-1/3) + 1 = 2/3

v3 = (-1/3) + 2 = 5/3

So the vector v is [-1/3, 2/3, 5/3]

Wait, let me verify if this satisfies all the original equations.

First, the dot product: 1*(-1/3) + 1*(2/3) + 1*(5/3) = (-1 + 2 + 5)/3 = 6/3 = 2. That's correct.

Cross product:

a × v = determinant as before. Let's compute each component.

i component: (1*(5/3) - 1*(2/3)) = 5/3 - 2/3 = 3/3 = 1. Correct.

j component: -(1*(5/3) - 1*(-1/3)) = -(5/3 + 1/3) = -6/3 = -2. Correct.

k component: (1*(2/3) - 1*(-1/3)) = 2/3 + 1/3 = 3/3 = 1. Correct.

So all equations are satisfied. Therefore, the vector v is [-1/3, 2/3, 5/3].

But let me check if there's another way to approach this, maybe using linear algebra. Since the cross product and dot product give two equations, and we have three variables, but the cross product gives three equations, but some are dependent.

Wait, but in this problem, we have four equations (dot product and cross product components) but with three variables. However, the cross product gives three equations which are not all independent. As we saw, Equation 2 was dependent on the others. So, in the end, we have three independent equations to solve for three variables, which we did. So the solution is unique.

Therefore, the answer is v = [-1/3, 2/3, 5/3]

**Final Answer**
The vector $\mathbf{v}$ is $\boxed{\begin{pmatrix} -\frac{1}{3} \\ \frac{2}{3} \\ \frac{5}{3} \end{pmatrix}}$.
Therefore, the vector $\mathbf{v}$ that satisfies the conditions is $\boxed{\begin{pmatrix} -\frac{1}{3} \\ \frac{2}{3} \\ \frac{5}{3} \end{pmatrix}}$.
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