Question

Equation of the plane containing the line $x+2y+3z-4=0=2x+y-z+5$ and perpendicular to the plane $5x+3y-6z+8=0$ is

• A. $33x+45y+50z=41$
• B. $33x+45y+50z=-41$
• C. $35x+45y+50z=41$
• D. $35x+45y+50z=-41$

Answer 1

The correct option is A $33x+45y+50z=41$

Equation of the plane containing the given line is
$P:(x+2y+3z-4)+\lambda (2x+y-z+5)=0$
$P:x(1+2\lambda )+y(2+\lambda )+z(3-\lambda )+(5\lambda -4)=0$

Since, $P$ is perpendicular to $5x+3y-6z+8=0.$
$\therefore 5\cdot (1+2\lambda )+3\cdot (2+\lambda )-6\cdot (3-\lambda )=0\Rightarrow \lambda =\frac{7}{19}$

The required equation of the plane is $33x+45y+50z=41$