The equation of the plane through the intersection of the planes $x+y+z=1$ and $2 x+3 y-z+4=0$ and parallel…
The equation of the plane through the intersection of the planes $x+y+z=1$ and $2 x+3 y-z+4=0$ and parallel to $x-a x i s$ is
- $3 y+z-6=0$
- $3 y-z+6=0$
- $y-3 z+6=0$
- $y+3 z-6=0$
Solution
Required equation is $(x+y+z-1)+\lambda(2 x+3 y-z+4)=0$
$\Rightarrow(1+2 \lambda) x+(1+3 \lambda) y+(1-\lambda) z+(4 \lambda-1)=0$
But plane is parallel to the $\mathrm{x}$-axis
$\begin{aligned}
& \Rightarrow \lambda=-\frac{1}{2} \\
& \Rightarrow-\frac{1}{2} y+\frac{3}{2} z-3=0 \\
& \Rightarrow y-3 z+6=0
\end{aligned}$
Asked in: MHT CET 2022 (10 Aug Shift 1)
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