Q. The equation of the plane passing through the lines $\frac{x - 4}{1} = \frac{y - 3}{1} = \frac{\mathcal{z} - 2}{2}$ and $\frac{x - 3}{1} = \frac{y - 2}{- 4} = \frac{\mathcal{z}}{5}$ is
A) $11x - y - 3\mathcal{z} = 35$
B) $11x + y - 3\mathcal{z} = 35$
C) $11x - y + 3\mathcal{z} = 35$
D) None of these
Explanation
(d)
Here, the required plane is
$a(x - 4) + b(y - 3) + c\left( \mathcal{z} - 2 \right) = 0$
Also $a + b + 2c = 0$ and $a - 4b + 5c = 0$
Solving, we have
$\frac{a}{5 + 8} = \frac{b}{2 - 5} = \frac{c}{- 4 - 1} = k$
$\frac{a}{13} = \frac{b}{- 3} = \frac{c}{- 5} = k$
Therefore, the required equation of plane is $- 13x + 3y + 5\mathcal{z} + 33 = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.