Question

What is the equation of the plane passing throught the line $3x+y-5z=2=x-2y+3z$ and perpendiculat to the plane $x-y+z=3$?

• A. $2x+3y+z=2$
• B. $3x+2y-z=2$
• C. $7(x-z)=6$
• D. no plane exists

Answer 1

Answer: (C)

$7(x-z)=6$

Equation of plane passing through line $3x+y-5z=2=x-2y+3z$ is given by,
$3x+y-5z-2+\lambda (x-2y+3z-2)=0$
$\Rightarrow (3+\lambda )x+(1-2\lambda )y+(3\lambda -5)z-2(1+\lambda )=0$
Also given that this plane is perpendicular to $x-y+z=3$
$\Rightarrow (3+\lambda )-(1-2\lambda )+(3\lambda -5)=0\Rightarrow \lambda =\frac{1}{2}$
Hence, required plane is
$7(x-z)=6$