Question

The perpendicular distance from the point $(2,-3,6)$ to the plane $3x-6y+$ $2z+10=0$ is

• A. $-\frac{46}{7}$
• B. $\frac{13}{7}$
• C. $\frac{46}{7}$
• D. $\frac{10}{7}$

Answer 1

The correct option is C $\frac{46}{7}$

Given equation of the plane is $ax+by+cz+d=3x-6y+$ $2z+10=0$ $a=3,b=-6,c=2,d=10$ and given point $(x_1,y_1,z_1)=(2,-3,6)$

$\therefore$ Perpendicular distance $p=$

$=\left|\frac{a{x}_1+b{y}_1+x{z}_1+d}{\sqrt{a^2+b^2+c^2}}\right|=\left|\frac{3\left(2\right)+(-6)(-3)+2\left(6\right)+10}{\sqrt{9+36+4}}\right|$

$=\left|\frac{6+18+12+10}{\sqrt{9+36+4}}\right|\therefore p=\frac{46}{\sqrt{49}}$