A line with positive direction cosines passes through the point  $A\left(1,2,-1\right)$and makes equal angles with the coordinate axes. The line meets the plane  $x+2y+z=5$at the point $B$. Then the point $B$and length of the line segment $\overline{AB}$respectively, are equal to

Equation of line passing through the point  $A\left(1,2,-1\right)$and making equal angles with axes, having positive direction ratios is $x-1=y-2=z+1$

General point on the line is $\left(k+1,k+2,k-1\right)$

Suppose that this point lies on the plane  $x+2y+z=5$

It gives 

            $\begin{array}{c}k+1+2k+4+k-1=5\\ 4k=1\\ k=\frac{1}{4}\end{array}$

Hence, the point of intersection of line and plane is $B\left(\frac{5}{4},\frac{9}{4},-\frac{3}{4}\right)$

The distance between  $A\left(1,2,-1\right),B\left(\frac{5}{4},\frac{9}{4},-\frac{3}{4}\right)$is  $AB=\sqrt{\frac{1}{16}+\frac{1}{16}+\frac{1}{16}}=\frac{\sqrt{3}}{4}$