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(B) Given points $P(3, -1, 2)$ and $Q(1, 2, -4)$.Direction vectors of lines $PR$ and $QS$ are $\vec{v_1} = 4\hat{i} - \hat{j} + 2\hat{k}$ and $\vec{v_

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$\sqrt{171}$

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(B) Given points $P(3, -1, 2)$ and $Q(1, 2, -4)$.Direction vectors of lines $PR$ and $QS$ are $\vec{v_1} = 4\hat{i} - \hat{j} + 2\hat{k}$ and $\vec{v_2} = -2\hat{i} + \hat{j} - 2\hat{k}$.The normal vector $\vec{n}$ to the plane containing $P, T, Q$ is $\vec{v_1} 	imes \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 4 & -1 & 2 \ -2 & 1 & -2 \end{vmatrix} = \hat{i}(2-2) - \hat{j}(-8+4) + \hat{k}(4-2) = 0\hat{i} + 4\hat{j} + 2\hat{k}$.Unit normal vector $\hat{n} = \frac{4\hat{j} + 2\hat{k}}{\sqrt{0^2 + 4^2 + 2^2}} = \frac{4\hat{j} + 2\hat{k}}{\sqrt{20}} = \frac{2\hat{j} + \hat{k}}{\sqrt{5}}$.For intersection point $T$,line $PT: \vec{r} = (3\hat{i} - \hat{j} + 2\hat{k}) + \lambda(4\hat{i} - \hat{j} + 2\hat{k})$ and line $QT: \vec{r} = (\hat{i} + 2\hat{j} - 4\hat{k}) + \mu(-2\hat{i} + \hat{j} - 2\hat{k})$.Equating coordinates: $3+4\lambda = 1-2\mu \Rightarrow 4\lambda + 2\mu = -2 \Rightarrow 2\lambda + \mu = -1$.$-1-\lambda = 2+\mu \Rightarrow \lambda + \mu = -3$.Subtracting equations: $\lambda = 2$,then $\mu = -5$.Point $T = (3+4(2), -1-(2), 2+2(2)) = (11, -3, 6)$.Vector $\vec{OA} = \vec{OT} \pm |\vec{TA}|\hat{n} = (11\hat{i} - 3\hat{j} + 6\hat{k}) \pm \sqrt{5} \left(\frac{2\hat{j} + \hat{k}}{\sqrt{5}}ight) = (11\hat{i} - 3\hat{j} + 6\hat{k}) \pm (2\hat{j} + \hat{k})$.Case $1$: $\vec{OA} = 11\hat{i} - \hat{j} + 7\hat{k} \Rightarrow |\vec{OA}| = \sqrt{121 + 1 + 49} = \sqrt{171}$.Case $2$: $\vec{OA} = 11\hat{i} - 5\hat{j} + 5\hat{k} \Rightarrow |\vec{OA}| = \sqrt{121 + 25 + 25} = \sqrt{171}$.

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