Equation of the plane passing through $(1,-1,2)$ and perpendicular to the planes $x+2 y-2 z=4$ and $3 x+2…
Equation of the plane passing through $(1,-1,2)$ and perpendicular to the planes $x+2 y-2 z=4$ and $3 x+2 y+z=6$ is
$6 x-7 y-4 z-5=0$
$6 x+7 y-4 z+5=0$
$6 x-7 y+4 z+5=0$
$6 x+7 y+4 z-5=0$
Solution
The equation of plane passing through $(1,-1,2)$ is $\mathrm{a}(x-1)+\mathrm{b}(y+1)+\mathrm{c}(\mathrm{z}-2)=0$
Since plane (i) is perpendicular to the planes $x+2 y-2 z=4$ and $3 x+2 y+z=6$
$\begin{array}{ll}
\therefore \quad & a+2 b-2 c=0 \\
& \text { and } 3 a+2 b+c=0 \\
& \Rightarrow \frac{a}{6}=\frac{b}{-7}=\frac{c}{-4}
\end{array}$
$\therefore \quad$ The equation of the required plane is
$\begin{aligned}
& 6(x-1)-7(y+1)-4(z-2)=0 \\
& \Rightarrow 6 x-7 y-4 z-5=0
\end{aligned}$