Question

The equation of the plane passing through $(2,-3,1)$ and is normal to the line joining the points $(3,4,-1)$ and $(2,-1,5)$ is given by

• A. $x+5y-6z+19=0$
• B. $x-5y+6z-19=0$
• C. $x+5y+6z+19=0$
• D. $x-5y-6z-19=0$

Answer 1

The correct option is A $x+5y-6z+19=0$

Equation of plane passing through $(2,-3,1)$ is
$a(x-2)+b(y+3)+c(z-1)=0$
and $D.r^{\prime }s$ of normal to plane $(a,b,c)=(3-2,4+1,-1-5)=(1,5,-6)$
So, equation of the plane is:
$1(x-2)+5(y+3)-6(z-1)=0\Rightarrow x+5y-6z+19=0$