Q. Let $A(1,\ 1,\ 1),\ B(2,\ 3,\ 5)$ and $C( - 1,\ 0,\ 2)$ be three points, then equation of a plane parallel to the plane $ABC$ which is at distance 2 is
A) $2x - 3y + \mathcal{z} + 2\sqrt{14} = 0$
B) $2x - 3y + \mathcal{z} - \sqrt{14} = 0$
C) $2x - 3y + \mathcal{z} + 2 = 0$
D) $2x - 3y + \mathcal{z} - 2 = 0$
Explanation
(a)
$A(1,\ 1,\ 1),\ B(2,\ 3,5),\ C( - 1,\ 0,\ 2)$ direction ratios of $AB$ are $< 1,\ 2,\ 4 >$
Direction ratios of $AC$ are $< - 2,\ - 1,\ 1 >$
Therefore, direction ratios of normal to plane $ABC$ are $< 2,\ - 3,\ 1 >$
As a result, equation of the plane $ABC$ is $2x - 3y + \mathcal{z} = 0$
Let the equation of the required plane is $2x - 3y + \mathcal{z} = k$, then $\left| \frac{k}{\sqrt{4 + 9 + 1}} \right| = 2$
$k = \pm 2\sqrt{14}$
Hence, equation of the required plane is $2x - 3y + \mathcal{z} + 2\sqrt{14} = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.