Let the line passing through the points (-1,2,1) and parallel to the line frac x -1 2 =frac y +1 3 =frac z 4 intersect the line x+2/3=y-3/2=z-4/1 at the point P . Then the distance of P from the point Q (4,-5,1) is :

Mathematics · 3D Geometry

JEE Main 2025 — 24 January, Morning Shift — Question 18

Let the line passing through the points $(-1,2,1)$ and parallel to the line $\frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}+1}{3}=\frac{\mathrm{z}}{4}$

intersect the line $\frac{x+2}{3}=\frac{y-3}{2}=\frac{z-4}{1}$ at the point $P$ . Then the distance of P from the point $\mathrm{Q}(4,-5,1)$ is :

Equation of line through point $(-1,2,1)$ is

$\Rightarrow \frac{\mathrm{x}+1}{2}=\frac{\mathrm{y}-2}{3}=\frac{\mathrm{z}-1}{4}=\lambda$

So, $\begin{array}{l}x=2 \lambda-1 , y=3 \lambda+2 , z=4 \lambda+1\end{array}.$

$\operatorname\rightarrow \frac{x+2}{3}=\frac{y-3}{2}=\frac{z-4}{1}=\mu($ Let $)$

So, $\begin{array}{l}x=3 \mu-2 , y=2 \mu+3 , z=\mu+4\end{array}.$

For intersection point ' P '

$\mathrm{x}=2 \lambda-1=3 \mu-2$

$y=3 \lambda+2=2 \mu+3 \quad\left[\begin{array}{l}\lambda=1 ,\mu=1\end{array}\right]$

$z=4 \lambda+1=\mu+4$

So, point $\mathrm{P}(\mathrm{x}, \mathrm{y}, \mathrm{z})=(1,5,5)$

$\& \mathrm{Q}(4,-5,1)$

$\therefore \mathrm{PQ}=\sqrt{9+100+16}$

$=\sqrt{125}=5 \sqrt{5}$

Option (4)

Answer key and solution verified before publishing.

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