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(D) Let $\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}$.Since $\vec{a}$ lies in the plane containing $\vec{b}$ and $\vec{c}$, $\vec{a}, \vec{b},$ and $\

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$264$

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(D) Let $\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}$.Since $\vec{a}$ lies in the plane containing $\vec{b}$ and $\vec{c}$, $\vec{a}, \vec{b},$ and $\vec{c}$ are coplanar. Thus, $\vec{a} \cdot (\vec{b} 	imes \vec{c}) = 0$.Calculating $\vec{b} 	imes \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 1 \ 2 & -1 & 1 \end{vmatrix} = \hat{i}(2+1) - \hat{j}(1-2) + \hat{k}(-1-4) = 3\hat{i} + \hat{j} - 5\hat{k}$.So, $(a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}) \cdot (3\hat{i} + \hat{j} - 5\hat{k}) = 0 \Rightarrow 3a_1 + a_2 - 5a_3 = 0 \dots (i)$.Since $\vec{a}$ is perpendicular to $\hat{i} + \hat{j} + 3\hat{k}$, we have $(a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}) \cdot (\hat{i} + \hat{j} + 3\hat{k}) = 0 \Rightarrow a_1 + a_2 + 3a_3 = 0 \dots (ii)$.The projection of $\vec{a}$ on $\vec{b}$ is $3\sqrt{6}$, so $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = 3\sqrt{6}$.$\frac{a_1 + 2a_2 + a_3}{\sqrt{1^2 + 2^2 + 1^2}} = 3\sqrt{6} \Rightarrow a_1 + 2a_2 + a_3 = 3\sqrt{6} \cdot \sqrt{6} = 18 \dots (iii)$.Solving equations $(i), (ii),$ and $(iii)$:Subtracting $(ii)$ from $(i)$: $2a_1 - 8a_3 = 0 \Rightarrow a_1 = 4a_3$.Substituting $a_1 = 4a_3$ into $(ii)$: $4a_3 + a_2 + 3a_3 = 0 \Rightarrow a_2 = -7a_3$.Substituting into $(iii)$: $4a_3 + 2(-7a_3) + a_3 = 18 \Rightarrow 4a_3 - 14a_3 + a_3 = 18 \Rightarrow -9a_3 = 18 \Rightarrow a_3 = -2$.Thus, $a_1 = -8$ and $a_2 = 14$.Therefore, $\vec{a} = -8\hat{i} + 14\hat{j} - 2\hat{k}$.$|\vec{a}|^2 = (-8)^2 + 14^2 + (-2)^2 = 64 + 196 + 4 = 264$.

Let $\vec{a}$ be a vector in the plane containing vectors $\vec{b}=\hat{i}+2 \hat{j}+\hat{k}$ and $\vec{c}=2 \hat{i}-\hat{j}+\hat{k}$. If $\vec{a}$ is perpendicular to $\hat{i}+\hat{j}+3 \hat{k}$ and its projection on $\vec{b}$ is $3 \sqrt{6}$, then $|\vec{a}|^2=$

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