If $\vec{a} = \hat{i} + \hat{k}$, $\vec{b} = \hat{j} - \hat{k}$ and $\vec{c} = \hat{i} + \hat{j} + \hat{k}$ such that $\vec{r} \times \vec{b} = \vec{c} \times \vec{b}$ and $\vec{r} \cdot \vec{a} = 0$, then $\vec{r}$ is:

Given three vectors:

$\vec{a} = \hat{i} + \hat{k}$

$\vec{b} = \hat{j} - \hat{k}$

$\vec{c} = \hat{i} + \hat{j} + \hat{k}$

And two conditions about the unknown vector $\vec{r}$:

$\vec{r} \times \vec{b} = \vec{c} \times \vec{b}$

$\vec{r} \cdot \vec{a} = 0$

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Starting with $\vec{r} \times \vec{b} = \vec{c} \times \vec{b}$:

$\vec{r} \times \vec{b} - \vec{c} \times \vec{b} = 0$

$(\vec{r} - \vec{c}) \times \vec{b} = 0$

When the cross product of two vectors equals zero, those vectors are parallel to each other.

So $\vec{r} - \vec{c}$ is parallel to $\vec{b}$.

This means:

$\vec{r} - \vec{c} = \lambda \vec{b}$

where $\lambda$ is some scalar.

Therefore:

$\vec{r} = \vec{c} + \lambda \vec{b}$

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Substituting the known vectors:

$\vec{r} = (\hat{i} + \hat{j} + \hat{k}) + \lambda(\hat{j} - \hat{k})$

$\vec{r} = \hat{i} + \hat{j} + \hat{k} + \lambda\hat{j} - \lambda\hat{k}$

$\vec{r} = \hat{i} + (1+\lambda)\hat{j} + (1-\lambda)\hat{k}$

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Using the condition $\vec{r} \cdot \vec{a} = 0$:

$[\hat{i} + (1+\lambda)\hat{j} + (1-\lambda)\hat{k}] \cdot [\hat{i} + \hat{k}] = 0$

$(1)(1) + (1+\lambda)(0) + (1-\lambda)(1) = 0$

$1 + 0 + 1 - \lambda = 0$

$2 - \lambda = 0$

$\lambda = 2$

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Substituting $\lambda = 2$:

$\vec{r} = \hat{i} + (1+2)\hat{j} + (1-2)\hat{k}$

$\vec{r} = \hat{i} + 3\hat{j} - \hat{k}$

Therefore, $\vec{r} = \hat{i} + 3\hat{j} - \hat{k}$