Q41 / 75 · JEE Main 2025 (29 Jan Shift 2) · Mathematics

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 $y z$-plane at the point $Q$, then the distance between the points $P$ and $Q$ is :

  • A
    $\sqrt{10}$
  • B
    $2 \sqrt{3}$
  • C
    2
  • Correct answer
    D
    3

Solution

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

Hence, the correct answer is option: 'D'.

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