Question

The equation of the plane which contains the line of intersection of planes $x+2y+3z-4=0$ and $2x+y-z+5=0$ which is perpandicular to the plane $5x+3y-6z+8=0$

• A. $3x+4y+5z-14=0$
• B. $33x+45y+50z-41=0$
• C. $37x+51y+49z-45=0$
• D. $7x+8y+15z-19=0$

Answer 1

The correct option is B $33x+45y+50z-41=0$

Given: $p_1:x+2y+3z-4=0,\cdots \left(i\right)p_2:2x+y-z+5=0\cdots \left(ii\right)p_3:5x+3y-6z+8=0\cdots \left(iii\right)$
The equation of the plane passing through the line of intersection of the planes $\left(i\right)\left(ii\right)$
$(x+2y+3z-4)+\lambda (2x+y-z+5)$
$(1+2\lambda )x+(2+\lambda )y+(3-\lambda )z-4+5\lambda =0\cdots \left(iv\right)$
This plane is perpandicular to $p_3$
So, $5(1+2\lambda )+3(2+\lambda )-6(3-\lambda )=0$
$19\lambda -7=0$
$\lambda =\frac{7}{19}$
putting in $\left(iv\right)$
$\Rightarrow (1+\frac{14}{19})x+(2+\frac{7}{19})y+(3-\frac{7}{19})z-4+5\times \frac{7}{19}=0$
$\Rightarrow 33x+45y+50z-41=0$