Question

lf the perpendicular distance from $(1,2,4)$ to the plane $2x+2y-z+k=0$ is $3$, then $k-4=$

• A. $0$
• B. $11$
• C. $3$
• D. $7$

Answer 1

Answer: (C)

$3$

Perpendicular distance of plane $2x+2y-z+k=0$ from the point $(1,2,4)$ is,

$=\left|\frac{2\left(1\right)+2\left(2\right)-1\left(4\right)+k}{\sqrt{2^2+2^2+1^2}}\right|=3$ .......(given)

$\Rightarrow \left|\frac{k+2}{3}\right|=3\Rightarrow k+2=\pm 9$

$\Rightarrow k=-11,7$

$\Rightarrow k-4=3$