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(D) Let $\bar{r} = a\hat{i} + b\hat{j} + c\hat{k}$.Since $\bar{r}$ is perpendicular to $\bar{q}$,we have $\bar{r} \cdot \bar{q} = 0$.This implies $a -

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$5(\hat{i}+\hat{j}+\hat{k})$

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(D) Let $\bar{r} = a\hat{i} + b\hat{j} + c\hat{k}$.Since $\bar{r}$ is perpendicular to $\bar{q}$,we have $\bar{r} \cdot \bar{q} = 0$.This implies $a - 2b + c = 0$ ... $(i)$.Since $\bar{r}$ is coplanar with $\bar{p}$ and $\bar{q}$,the scalar triple product $[\bar{p} \ \bar{q} \ \bar{r}] = 0$.$\begin{vmatrix} 1 & 1 & 1 \ 1 & -2 & 1 \ a & b & c \end{vmatrix} = 0$.Expanding the determinant: $1(-2c - b) - 1(c - a) + 1(b + 2a) = 0$.$-2c - b - c + a + b + 2a = 0 \Rightarrow 3a - 3c = 0 \Rightarrow a = c$ ... $(ii)$.Substituting $(ii)$ into $(i)$: $a - 2b + a = 0 \Rightarrow 2a = 2b \Rightarrow a = b$.Thus,$\bar{r} = a(\hat{i} + \hat{j} + \hat{k})$.The unit vector in the direction of $\bar{r}$ is $\frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}$.The required vector has magnitude $5\sqrt{3}$,so $	ext{Vector} = 5\sqrt{3} 	imes \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}} = 5(\hat{i} + \hat{j} + \hat{k})$.

If $\overline{p}=\hat{i}+\hat{j}+\hat{k}$ and $\overline{q}=\hat{i}-2 \hat{j}+\hat{k}$. Then a vector of magnitude $5 \sqrt{3}$ units perpendicular to the vector $\overline{q}$ and coplanar with $\overline{p}$ and $\overline{q}$ is

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