JEE MAIN - Mathematics (2025 - 29th January Evening Shift - No. 18)
Let the line $\vec{r}=(-\hat{i}+\hat{j}-2 \hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k}), \lambda \in \mathbf{R}$, intersect the line L at Q . Then $2(\mathrm{PQ})^2$ is equal to :
Explanation
$$\begin{aligned} & \mathrm{L}: \frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}+1}{-1}=\frac{\mathrm{z}-2}{2}=\mu \\ & \mathrm{P}(\mu+1,-\mu-1,2 \mu+2) \\ & \overrightarrow{\mathrm{AP}} \cdot \overrightarrow{\mathrm{~d}}=0 \Rightarrow(\mu,-\mu-3,2 \mu) \cdot(1,-1,2)=0 \\ & \Rightarrow \mu+\mu+3+4 \mu=0 \Rightarrow \mu=-\frac{1}{2} \\ & \therefore \mathrm{P}\left(\frac{-1}{2}+1,+\frac{1}{2}-1,2\left(\frac{-1}{2}\right)+2\right) \\ & \mathrm{P}\left(\frac{1}{2}, \frac{-1}{2}, 1\right) \end{aligned}$$
$$\begin{array}{c|c|c} \mu+1=-1+\lambda & -\mu-1=1-\lambda & 2 \mu+2=-2+\lambda \\ \mu=\lambda-2 & \mu=\lambda-2 & \downarrow \end{array}$$
$$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\begin{aligned} & 2(\lambda-2)+2=-2+\lambda \\ & 2 \lambda-4+2=-2+\lambda \end{aligned}$$
$$\begin{aligned} & \therefore \quad \mu=-2, \lambda=0 \\ & \therefore \quad \mathrm{Q} \equiv(-1,1-2) \\ & \mathrm{P}\left(\frac{1}{2}, \frac{-1}{2}, 1\right) \text { and } \mathrm{Q}(-1,1,-2) \\ & \mathrm{PQ}=\sqrt{\left(\frac{1}{2}+1\right)^2+\left(\frac{-1}{2}-1\right)^2+(1+2)^2} \\ & =\sqrt{\frac{9}{4}+\frac{9}{4}+9}=\sqrt{\frac{54}{4}} \\ & \therefore \quad 2(\mathrm{PQ})^2=2\left(\frac{54}{4}\right)=27 \end{aligned}$$
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