Question

Find the distance from the plane passing through $(1,3,2)$, $(-5,0,2)$, $(1,1,-4)$ to the point $(2,3,4)$.

Answer 1

$A(1,3,2),B(-5,0,2),C(1,1,-4$

normal vector will be $AB\times BC=18i-36j+12k=\alpha (3i-6j+2k)$

Equation of plane $3x-6y+2z=c$ will satisfy point $C(1,1,-4)$

so, $3\left(1\right)-6\left(1\right)+2(-4)=c$

$c=-11$

equation of plane will be $3x-6y+2z+11=0$

distance of $(2,3,4)$ from this plane will be $\frac{3\left(2\right)-6\left(3\right)+2\left(4\right)+11}{\sqrt{3^2+(-6{)}^2+2^2}}=\frac{7}{7}=1$unit