Mathematics · 3D Geometry

JEE Main 2025 — 4 April, Evening Shift — Question 30

Let $A$ be the point of intersection of the lines $L_{1}: \frac{x-7}{1}=\frac{y-5}{0}=\frac{z-3}{-1}$ and

$L_{2}: \frac{x-1}{3}=\frac{y+3}{4}=\frac{z+7}{5}$ . Let $B$ and $C$ be the points on the lines $L_{1}$ and $L_{2}$ respectively such that

$A B=A C=\sqrt{15}$ . Then the square of the area of the triangle $A B C$ is:

Answer: D

$L_{1}: \frac{x-7}{1}=\frac{y-5}{0}=\frac{z-3}{-1} ; L_{2}: \frac{x-1}{3}=\frac{y+3}{4}=\frac{z+7}{5}$

$\cos \theta=\left|\frac{3+0-5}{\sqrt{2} \times \sqrt{50}}\right|$

$=\frac{2}{10}=\frac{1}{5}$

$\therefore \sin \theta=\frac{2 \sqrt{6}}{5}$

Area $=\frac{1}{2} a b \sin \theta$

$=\frac{1}{2} \times \sqrt{15} \times \sqrt{15} \times \frac{2 \sqrt{6}}{5}$

$=3 \sqrt{6}$

$(\text { Area })^{2}=9 \times 6=54$

Answer key and solution verified before publishing.

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