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(B) Let $\vec{x} = a\hat{i} + b\hat{j} + c\hat{k}$.Since $\vec{x}$ is a unit vector,$a^2 + b^2 + c^2 = 1$.The cross product is given by:$\vec{x} 	ime

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$-\frac{1}{3}(2\hat{i} - \hat{j} + 2\hat{k})$

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(B) Let $\vec{x} = a\hat{i} + b\hat{j} + c\hat{k}$.Since $\vec{x}$ is a unit vector,$a^2 + b^2 + c^2 = 1$.The cross product is given by:$\vec{x} 	imes (\hat{i} - 2\hat{j} + \hat{k}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ a & b & c \ 1 & -2 & 1 \end{vmatrix} = (b+2c)\hat{i} - (a-c)\hat{j} + (-2a-b)\hat{k}$.Equating this to $-\hat{i} + \hat{k}$,we get:$b + 2c = -1$,$a - c = 0$,and $-2a - b = 1$.From $a = c$,substituting into the other equations: $b + 2a = -1$ and $-2a - b = 1$,which are the same equation.Using $a^2 + b^2 + c^2 = 1$,we get $a^2 + (-1-2a)^2 + a^2 = 1$.$a^2 + 1 + 4a + 4a^2 + a^2 = 1 \Rightarrow 6a^2 + 4a = 0$.So,$2a(3a + 2) = 0$,which gives $a = 0$ or $a = -\frac{2}{3}$.If $a = 0$,then $c = 0$ and $b = -1$,so $\vec{x} = -\hat{j}$.If $a = -\frac{2}{3}$,then $c = -\frac{2}{3}$ and $b = -1 - 2(-\frac{2}{3}) = -1 + \frac{4}{3} = \frac{1}{3}$.Thus,$\vec{x} = -\frac{2}{3}\hat{i} + \frac{1}{3}\hat{j} - \frac{2}{3}\hat{k} = -\frac{1}{3}(2\hat{i} - \hat{j} + 2\hat{k})$.

If $\vec{x}$ is a unit vector such that $\vec{x} \times (\hat{i} - 2\hat{j} + \hat{k}) = -\hat{i} + \hat{k}$,then $\vec{x}$ is:

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