The line drawn $(4,-1,2)$ and $(-3,2,3)$ meets the plane at right angles at the point $(-10,5,4)$ then the…
The line drawn $(4,-1,2)$ and $(-3,2,3)$ meets the plane at right angles at the point $(-10,5,4)$ then the equation of plane is
$2 x-y-z+29=0$
$7 x-3 y-z+89=0$
$x-y+z+11=0$
$x+y+z+1=0$
Solution
d.r's of normal to the plane are $ < 4-(-3),-1-2$, $2-3>^0 < 7,-3,-1>$
and it passes through $(-10,5,4)$. Hence the required equation is
$\begin{aligned}
& 7(x-(-10))-3(y-5)-1(z-4)=0 \\
& \Rightarrow 7 x-3 y-z+89=0
\end{aligned}$