Question

The points $A(-1,3,0),B(2,2,1)$ and $C(1,1,3)$ determine a plane. The distance of the plane $ABC$ from the point $D(5,7,8)$ is

• A. $\sqrt{66}$
• B. $\sqrt{71}$
• C. $\sqrt{73}$
• D. $\sqrt{76}$

Answer 1

The correct option is A $\sqrt{66}$

Equation of plane passing through $A(-1,3,0),B(2,2,1),C(1,1,3)$ will have direction$AB\times BC=-i-7j-4k$

so plane $x+7y+4z=c$ will contain $C(1,1,3)$ so itwill satisfy plane'equation

$1+7\left(1\right)+4\left(3\right)=c$

$c=20$

hence equation of plane will be $x+7y+4z=20$

distance of $(5,7,8)$ from the plane will be $\frac{1\left(5\right)+7\left(7\right)+4\left(8\right)-20}{\sqrt{1^2+7^2+4^2}}=\frac{66}{\sqrt{66}}=\sqrt{66}$unit