Let bar{a} = 2bar{i} - bar{j} + bar{k},bar{b} | vedclass

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(C) Since $\bar{r}$ lies in the plane of $\bar{b}$ and $\bar{c}$,we can write $\bar{r} = \bar{b} + m\bar{c}$ for some scalar $m \in \mathbb{R}$.Substi

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$-2\bar{i} - \bar{j} + 5\bar{k}$

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(C) Since $\bar{r}$ lies in the plane of $\bar{b}$ and $\bar{c}$,we can write $\bar{r} = \bar{b} + m\bar{c}$ for some scalar $m \in \mathbb{R}$.Substituting the given vectors: $\bar{r} = (\bar{i} + 2\bar{j} - \bar{k}) + m(\bar{i} + \bar{j} - 2\bar{k}) = (1+m)\bar{i} + (2+m)\bar{j} - (1+2m)\bar{k}$.The magnitude of the projection of $\bar{r}$ on $\bar{a}$ is given by $\frac{|\bar{r} \cdot \bar{a}|}{|\bar{a}|} = \sqrt{\frac{2}{3}}$.Calculating $|\bar{a}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{6}$.Thus,$\frac{|2(1+m) - (2+m) - (1+2m)|}{\sqrt{6}} = \sqrt{\frac{2}{3}} = \frac{\sqrt{2}}{\sqrt{3}} = \frac{\sqrt{6}}{3}$.$|2 + 2m - 2 - m - 1 - 2m| = \sqrt{6} \cdot \frac{\sqrt{6}}{3} = 2$.$|-m - 1| = 2$,which implies $m+1 = 2$ or $m+1 = -2$.Case $1$: $m = 1$. Then $\bar{r} = (1+1)\bar{i} + (2+1)\bar{j} - (1+2)\bar{k} = 2\bar{i} + 3\bar{j} - 3\bar{k}$.Case $2$: $m = -3$. Then $\bar{r} = (1-3)\bar{i} + (2-3)\bar{j} - (1-6)\bar{k} = -2\bar{i} - \bar{j} + 5\bar{k}$.Comparing with the given options,the correct vector is $-2\bar{i} - \bar{j} + 5\bar{k}$.

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