Let a straight line L pass through the point P(2,–1,3) and be perpendicular to the lines
$\frac{x - 1}{2} = \frac{y + 1}{1} = \frac{z - 3}{- 2} and \frac{x - 3}{1} = \frac{y - 2}{3} = \frac{z + 2}{4} .$ If the line L intersects the yz-plane at the point Q, then the distance between the points P and Q is :

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$\sqrt{10}$

2 $\sqrt{3}$

answer is C.

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Detailed Solution

Vector parallel to 'L'
$\begin{array}{l} = |\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & - 2 \\ 1 & 3 & 4 \end{array}| = 10 \hat{i} - 10 \hat{j} + 5 \hat{k} \\ = 5 (2 \hat{i} - 2 \hat{j} + \hat{k}) \end{array}$
Equation of 'L'
$\begin{array}{l} \frac{x - 2}{2} = \frac{y + 1}{- 2} = \frac{z - 3}{1} = λ (say) \\ Let Q (2 λ + 2, - 2 λ - 1, λ + 3) \\ \Rightarrow 2 λ + 2 = 0 \Rightarrow λ = - 1 \end{array}$
$\Rightarrow$ Q(0, 1, 2)
d(P, Q) = 3