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 point on the lines $L_1$ and $L_2$ respectively such that $\mathrm{AB}=\mathrm{AC}=\sqrt{15}$. Then the square of the area of the triangle ABC is :

• Correct answer
  A

  $54$
• B

  $63$
• C

  $57$
• D

  $60$

Solution

Angle between both lines
$\begin{aligned}<br/>& \cos \theta=\left|\frac{3+0-5}{\sqrt{2} \sqrt{50}}\right| \\ & \sin \theta=\frac{2}{10}=\frac{1}{5} \\ & \sin \theta=\frac{\sqrt{24}}{5} \\ & \text { area }=\frac{1}{2} a b \sin \theta \\ & \frac{1}{2} \sqrt{15} \sqrt{15} \frac{\sqrt{24}}{5}<br/>\end{aligned}$
square of area $\frac{15.15 .24}{4.25}$

option (1)

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