The equation of the plane which passes through $(2,-3,1)$ and is normal to the line joining the points $(3,4…
The equation of the plane which passes through $(2,-3,1)$ and is normal to the line joining the points $(3,4,-1)$ and $(2,-1,5)$ is given by
$x+5 y-6 z+19=0$
$x-5 y+6 z-23=0$
$x+5 y+6 z+7=0$
$x-5 y-6 z-11=0$
Solution
The required plane passes through $(2,-3,1)$.
It is normal to the line having d.r.s. $(1,5,-6)$.
$\therefore \mathrm{x}+5 \mathrm{y}-6 \mathrm{z}=\mathrm{k} \Rightarrow 2+5(-3)-6(1)=1$ i.e. $\mathrm{k}=-19$
Hence equation of plane is $x+5 y-6 z+19=0$