Question

Find the equation of the plane passing through $(-2,1,3)$ and having $(3,-5,4)$ as $d.r^{\prime }s$ of its normal.

Answer 1

equation of plane having $(3,-5,4)$ as d.r's is $3x-5y+4z=c$

This plane pass through$(-2,1,3)$ so the point will satisfy the equation

$3(-2)-5\left(1\right)+4\left(3\right)=c$

$c=1$

So, equation of plane is $3x-5y+4z=1$