3D Geometry 3 Question 34
####34. The equation of the plane passing through the point $(1,1,1)$ and perpendicular to the planes $2 x+y-2 z=5$ and $3 x-6 y-2 z=7$ is
(2017 Adv.)
(a) $14 x+2 y-15 z=1$
(b) $-14 x+2 y+15 z=3$
(c) $14 x-2 y+15 z=27$
(d) $14 x+2 y+15 z=31$
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Answer:
Correct Answer: 34. (d)
Solution:
- Let the equation of plane be $a x+b y+c z=1$. Then
$ \begin{array}{r} a+b+c=1 \\ 2 a+b-2 c=0 \\ 3 a-6 b-2 c=0 \Rightarrow a=7 b, c=\frac{15 b}{2} \\ b=\frac{2}{31}, a=\frac{14}{31}, c=\frac{15}{31} \\ \therefore 14 x+2 y+15 z=31 \end{array} $
