Question

The equation of the plane through intersection of planes $x+2y+3z=4$ and $2x+y-z=-5,$ and perpendicular to the plane $5x+3y+6z+8=0$ is

• A. $7x-2y+3z+81=0$
• B. $23x+14y-9z+48=0$
• C. $51x+15y-50z+173=0$
• D. $5x-17y+2z+23=0$

Answer 1

The correct option is C $51x+15y-50z+173=0$

Equation of plane through intersection of the two planes is,
$P:(x+2y+3z-4)+\lambda (2x+y-z+5)=0\dots \left(1\right)$
${\overrightarrow{n}}_P=(1+2\lambda )\hat{i}+(2+\lambda )\hat{j}+(3-\lambda )\hat{k}$
Direction ratios of its normal are $2\lambda +1,\lambda +2,3-\lambda$

Again, consider the plane $5x+3y+6z+8=0\dots \left(2\right)$
Direction ratios of its normal are $5,3,6$

Since plane $\left(1\right)$ is perpendicular to plane $\left(2\right),$
$n_P\cdot \overrightarrow{n}=0\Rightarrow 5(1+2\lambda )+3(2+\lambda )+6(3-\lambda )=0$
$\Rightarrow 7\lambda +29=0\Rightarrow \lambda =-\frac{29}{7}$
Putting $\lambda =-\frac{29}{7}$ in equation $\left(1\right),$ we get
$(x+2y+3z-4)-\frac{29}{7}(2x+y-z+5)=0\Rightarrow 7(x+2y+3z-4)-29(2x+y-z+5)=0\Rightarrow 7x+14y+21z-28-58x-29y+29z-145=0\Rightarrow 51x+15y-50z+173=0$