Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Equation of plane passing through $(2,2,1)$ is $\begin{array}{ll}& \mathrm{a}(\mathrm{x}-2)+\mathrm{b}(\mathrm{y}-2)+\mathrm{c}(\mathrm{z}-1)=0\\ & \text{ Since, above plane is perpendicular to }\\ & 3\mathrm{x}+2\mathrm{y}+4\mathrm{z}+1=0\\ & \text{ and }2\mathrm{x}+\mathrm{y}+3\mathrm{z}+2=0\\ \therefore & 3\mathrm{a}+2\mathrm{b}+4\mathrm{c}=0\\ & \text{ and }2\mathrm{a}+\mathrm{b}+3\mathrm{c}=0\\ & [\because \text{ for perpendicular, }{\mathrm{a}}_1{\mathrm{a}}_2\\ & +{\mathrm{b}}_1{\mathrm{b}}_2+{\mathrm{c}}_1{\mathrm{c}}_2=0]\end{array}$ On multiplying eq. (iii) by 2 , we get $4a+2b+6c=0$ On subtracting eq. (iv) from eq. (ii), we get $\Rightarrow c=\frac{-a}{2}$ 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), $\begin{array}{rl}& \text{ we get }a(x-2)-\frac{a}{2}(y-2)-\frac{a}{2}(z-1)=0\\ \Rightarrow & \frac{a}{2}[2(x-2)-(y-2)-(z-1)]=0\\ \Rightarrow & 2x-4-y+2-z+1=0\\ \Rightarrow & 2x-y-z-1=0\end{array}$