Question

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

• A. $2x+3y+z=2$
• B. $3x+2y-z=2$
• C. $7(x-z)=6$
• D. $19x-8y-27z=14$

Answer 1

Answer: (D)

$19x-8y-27z=14$

Equation of the plane passing through the line $3x-y-5z=2=x-2y+3z$ is

$3x-y-5z-2+k(x-2y+3z-2)=0$

$\Rightarrow (3+k)x-(1+2k)y+(-5+3k)z=2+2k$

Since, it is perpendicular to $x-y+z=3$

Therefore, $\left(\right(3+k)i-(1+2k)j+(-5+3k\left)k\right).(i-j+k)=0$

$\Rightarrow 3+k+1+2k-5+3k=0$ $\Rightarrow k=\frac{1}{6}$

Therefore, desired plane is $\Rightarrow (3+\frac{1}{6})x-(1+2\left(\frac{1}{6}\right))y+(-5+3\left(\frac{1}{6}\right))z=2+2\left(\frac{1}{6}\right)$

$\Rightarrow 19x-8y-27=14$