Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

The perpendicular distance from a point $(x_1,y_1,z_1)$ to the plane $ax+by+cz+d=0$ is given by $p=\frac{|a{x}_1+b{y}_1+c{z}_1+d|}{\sqrt{a^2+b^2+c^2}}$.

Substituting the given values $(x_1,y_1,z_1)=\left(1,\frac{3}{2},2\right)$ and the plane equation $2x-2y+4z+17=0$:

$p=\frac{\left|2(1)-2\left(\frac{3}{2}\right)+4(2)+17\right|}{\sqrt{2^2+(-2{)}^2+4^2}}$

$p=\frac{|2-3+8+17|}{\sqrt{4+4+16}}$

$p=\frac{24}{\sqrt{24}}$

$p=\sqrt{24}=2\sqrt{6}$ units.

Answer: $2\sqrt{6}$ units