A line passing through $A\left(1,2,3\right)$ and having direction ratios $〈3,4,5〉$meets a plane $x+2y-3z=5$  at $B$ then distance $AB$ is equal to

The equation of the line passing through the point $A\left(1,2,3\right)$ and having direction ratios $〈\mathrm{3.4.5}〉$  is $\frac{x-1}{3}=\frac{y-2}{4}=\frac{z-3}{5}$

General point on the line is $\left(3k+1,4k+2,5k+3\right)$

If it lies on the plane$x+2y-3z=5$

It gives 

$\begin{array}{c}3k+1+8k+4-15k-9=5\\ -4k=9\\ k=-\frac{9}{4}\end{array}$            

Hence the point $B\left(-\frac{23}{4},-7,-\frac{33}{4}\right)$

Therefore, the distance is  $AB=\sqrt{{\left(\frac{27}{4}\right)}^2+81+{\left(\frac{45}{4}\right)}^2}=\frac{\sqrt{4050}}{4}=\boxed{\frac{45\sqrt{2}}{4}}$