Question

Let two points $A=(3,1,2)$ and $B=(1,2,-4).$ Then the distance of the point $C(-1,1,1)$ from the plane passing through $B$ and perpendicular to $AB$ is $\frac{27}{\sqrt{\alpha }}$. Then the value of $\alpha$ is

Answer 1

Given points : $A=(3,1,2)$ and $B=(1,2,-4).$
Equation of plane is passing through $B$:
$a(x-1)+b(y-2)+c(z+4)=0\cdots \left(i\right)$
and $D.r^{\prime }s$ of normal is $(a,b,c)=D.r^{\prime }s$ of lines $AB=(-2,1,-6)$
So plane is $-2(x-1)+1(y-2)-6(z+4)=0$
$\Rightarrow -2x+y-6z-24=0$
So, perpendicular distance from $C$ is
$=\left|\frac{-2(-1)+1\left(1\right)-6\left(1\right)-24}{\sqrt{2^2+1^1+6^2}}\right|=\frac{27}{\sqrt{41}}$
So, $\alpha =41$