Question

The line $\frac{x - 1}{\lambda }=\frac{y}{2}=\frac{z + 3}{- 12}$ makes an isosceles triangle with the planes $x+2y-3z+5=0$ and $2x+3y-z+2=0,$ then value of $\lambda $ can be

• A. $5$
• B. $11$
• C. $22$
• D. $9$

Answer 1

Answer: (C)

Line makes equal angles with both planes and hence equal angles with normals of both planes
$\frac{\lambda + 4 + 36}{\sqrt{\lambda ^{2} + 4 + 144} \sqrt{1^{2} + 2^{2} + 3^{2}}}=\frac{2 \lambda + 6 + 12}{\sqrt{\lambda ^{2} + 4 + 144} \sqrt{1^{2} + 2^{2} + 3^{2}}}\therefore \lambda =22$