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JEE Mains · Maths · STD 12 - 11. three dimension geometry

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 :

  1. A \(54\)
  2. B \(63\)
  3. C \(57\)
  4. D \(60\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(54\)

Step-by-step Solution

Detailed explanation

Angle between both lines…