Question

The distance of the point $A(1,2,-1)$ from the plane $x-2y+4z=10$ is $k$ then $21k^2$ is

Answer 1

Given plane equation is $x-2y+4z=10,A(1,2,-1)\text{Distance}=\frac{|{lx}_1+{my}_1+{nz}_1+d|}{\sqrt{l^2+m^2+n^2}}\text{Here}l=1,m=-2,n=4,d=-10,x_1=1,y_1=2,z_1=-1$
Distance $=\frac{|1\left(1\right)-2\left(2\right)+4(-1)-10|}{\sqrt{1+4+16}}$
$\therefore \text{Distance}=\frac{|17|}{\sqrt{21}}$