The distance between the point $\left(-1,-5,-10\right)$ and the point of intersection of the line $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}$ with the plane $x-y+z=5$ is $13t,$ then $t$ equals to

Any point on the line is $\left(3r+2, 4r –1, 12r+2\right)$
If it lies on the given plane, then $\left(3r+2\right)-\left(4r-1\right)+\left(12r+2\right)=5$
$\Rightarrow 11r=0 \Rightarrow r=0$
$\therefore$ Co-ordinates of the point of intersection of the given line and given plane are $\left(2, -1, 2\right)$ and thus the required distance is $\sqrt{{\left(2+1\right)}^2+{\left(-1+5\right)}^2+{\left(2+10\right)}^2}=13$

$\Rightarrow 13=13t \Rightarrow t=1$