The equation of the plane, passing through the point $(1,1,1)$ and perpendicular to the planes $2 x+y-2 z=5$…

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
  1. $14 x+2 y-15 z=1$
  2. $14 x-2 y+15 z=27$
  3. $14 x+2 y+15 z=31$
  4. $-14 x+2 y+15 z=3$

Solution

The equation of plane passing through $(1,1,1)$ is $a(x-1)+b(y-1)+c(z-1)=0$
Since plane (i) is perpendicular to the planes $2 x+y-2 z=5 \text { and } 3 x-6 y-2 z=7$ $\therefore \quad 2 \mathrm{a}+\mathrm{b}-2 \mathrm{c}=5$...(i) $3 a-6 b-2 c=7...(ii)$
On solving (i), (ii) and (iii), we get $a=14, b=2, c=15$
Substituting the values of $a, b, c$ in (i), we get $\begin{aligned} & 14(x-1)+2(y-1)+15(z-1)=0 \\ & \Rightarrow 14 x+2 y+15 z=31 \end{aligned}$

Asked in: MHT CET 2024 (09 May Shift 1)

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