JEE MAIN - Mathematics (2025 - 4th April Evening Shift - No. 14)
Let A be the point of intersection of the lines $\mathrm{L}_1: \frac{x-7}{1}=\frac{y-5}{0}=\frac{z-3}{-1}$ and $\mathrm{L}_2: \frac{x-1}{3}=\frac{y+3}{4}=\frac{z+7}{5}$. Let B and C be the points on the lines $\mathrm{L}_1$ and $\mathrm{L}_2$ respectively such that $A B=A C=\sqrt{15}$. Then the square of the area of the triangle $A B C$ is :
63
57
60
54
Explanation
$L_1: \frac{x-7}{1}=\frac{y-5}{0}=\frac{z-3}{-1} ; L_2: \frac{x-1}{3}=\frac{y+3}{4}=\frac{z+7}{5}$
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$$\begin{aligned} & \cos \theta=\left|\frac{3+0-5}{\sqrt{2} \times \sqrt{50}}\right| \\ & =\frac{2}{10}=\frac{1}{5} \\ & \therefore \sin \theta=\frac{2 \sqrt{6}}{5} \\ & \text { Area }=\frac{1}{2} a b \sin \theta \\ & =\frac{1}{2} \times \sqrt{15} \times \sqrt{15} \times \frac{2 \sqrt{6}}{5} \\ & =3 \sqrt{6} \\ & (\text { Area })^2=9 \times 6=54 \end{aligned}$$
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