A plane bisects the line joining $A(-1,2,1)$ and
$B(1, 3, 4)$ and is perpendicular to AB. The distance
of the point;.$(4, 1, 5)$ from the plane is

If $B$ is the image of $A,$ then the locus of a point
$P(x, y, z)$ which is equidistant from $A$ and $B$ is a plane 
bisecting $AB$ perpendicularly.

its iquation is given by $PA=PB\text{. }$

$\begin{array}{l}(x+1{)}^2+(y-2{)}^2+(z-1{)}^2=(x-1{)}^2+(y-3{)}^2+(z-4{)}^2\\ \Rightarrow 4x+2y+6z=20\text{ or }2x+y+3z=10\end{array}$

The distance of $(4, 1, 5)$ from it is

$\frac{8+1+15-10}{\sqrt{4+1+9}}=\frac{14}{\sqrt{14}}=\sqrt{14}$