Question

If the line $\frac{x-1}{2}=\frac{y+3}{1}=\frac{z-5}{-1}$ is parallel to the plane $px+3y-z+5=0$, then the value of $p$

• A. $2$
• B. $-2$
• C. $\frac{1}{2}$
• D. $\frac{1}{3}$

Answer 1

The correct option is B $-2$

line $11$ plane

$\therefore$ line $\perp$ normal to plane

$\therefore \left(2\right)\left(P\right)+\left(1\right)\left(3\right)+(-1)(-1)=0$

$\therefore 2p+3+1=0$

$\therefore P=-2$