Q75 / 120 · VITEEE 2015 · Mathematics
If a plane passing through the point $(2,2,1)$ and is perpendicular to the planes $3 x+2 y+4 z+1=0$ and $2 x+y+3 z+2=0$. Then, the equation of the plane is
- Correct answerA$2 x-y-z-1=0$
- B$2 x+3 y+z-1=0$
- C$2 x+y+z+3=0$
- D$x-y+z-1=0$
Solution
Equation of plane passing through $(2,2,1)$ is
$\begin{array}{ll} & \mathrm{a}(\mathrm{x}-2)+\mathrm{b}(\mathrm{y}-2)+\mathrm{c}(\mathrm{z}-1)=0 \\ & \text { Since, above plane is perpendicular to } \\ & 3 \mathrm{x}+2 \mathrm{y}+4 \mathrm{z}+1=0 \\ & \text { and } 2 \mathrm{x}+\mathrm{y}+3 \mathrm{z}+2=0 \\ \therefore & 3 \mathrm{a}+2 \mathrm{~b}+4 \mathrm{c}=0 \\ & \text { and } 2 \mathrm{a}+\mathrm{b}+3 \mathrm{c}=0 \\ & {\left[\because \text { for perpendicular, } \mathrm{a}_{1} \mathrm{a}_{2}\right.} \\ & \left.\quad+\mathrm{b}_{1} \mathrm{~b}_{2}+\mathrm{c}_{1} \mathrm{c}_{2}=0\right]\end{array}$
On multiplying eq. (iii) by 2 , we get $4 a+2 b+6 c=0$
On subtracting eq. (iv) from eq. (ii), we get
$\Rightarrow \quad c=\frac{-a}{2}$
On putting $\mathrm{c}=\frac{-\mathrm{a}}{2}$ in eq. (iii), we get $\mathrm{b}$
$=\frac{-\mathrm{a}}{2}$
On putting $b=\frac{-a}{2}$ and $c=\frac{-a}{2}$ in eq. (i),
$\begin{aligned}
& \text { we get } a(x-2)-\frac{a}{2}(y-2)-\frac{a}{2}(z-1)=0 \\
\Rightarrow & \frac{a}{2}[2(x-2)-(y-2)-(z-1)]=0 \\
\Rightarrow & 2 x-4-y+2-z+1=0 \\
\Rightarrow & 2 x-y-z-1=0
\end{aligned}$
Hence, the correct answer is option: 'A'.