It is known that the equation of the line passing through the points $\left({\mathrm{x}}_1,{\mathrm{y}}_1,{\mathrm{z}}_1\right)$ and $\left({\mathrm{x}}_2,{\mathrm{y}}_2,{\mathrm{z}}_2\right)$ is
$\frac{\mathrm{x}-{\mathrm{x}}_1}{{\mathrm{x}}_2-{\mathrm{x}}_1}=\frac{\mathrm{y}-{\mathrm{y}}_1}{{\mathrm{y}}_2-{\mathrm{y}}_1}=\frac{\mathrm{z}-{\mathrm{z}}_1}{{\mathrm{z}}_2-{\mathrm{z}}_1}$
So here we have,

The line passing through the points, $\left(3,-4,-5\right)$ and $\left(2,-3,1\right)$ is given by,
$\frac{\mathit{x}-3}{2-3}=\frac{\mathit{y}+4}{-3+4}=\frac{\mathit{z}+5}{1+5}$
Let $\frac{\mathrm{x}-3}{-1}=\frac{\mathrm{y}+4}{1}=\frac{\mathrm{z}+5}{6}=\mathrm{k}$
Now,
$\Rightarrow \frac{\mathrm{x}-3}{-1}=\mathrm{k}$
$\therefore \mathrm{x}=-\mathrm{k}+3$
$\Rightarrow \frac{\mathrm{y}+4}{1}=\mathrm{k}$
$\therefore \mathrm{y}=\mathrm{k}-4$
$\Rightarrow \frac{\mathrm{z}+5}{6}=\mathrm{k}$
$\therefore \mathrm{z}=6\mathrm{k}-5$
The co-ordinates of the point of intersection of the given line and plane is, $\left(-\mathrm{k}+3,\mathrm{k}-4,6\mathrm{k}-5\right)$
If is lies on the plane $2\mathrm{x}+\mathrm{y}+\mathrm{z}=7$, then,
$\Rightarrow 2\left(-\mathrm{k}+3\right)+\mathrm{k}-4+6\mathrm{k}-5=7$
$\Rightarrow 5\mathrm{k}=10$
$\therefore \mathrm{k}=2$
Substituting k = 2 in ( 1 ) we get,
The co-ordinates of the point of intersection of the given line and plane are
$\left(-\mathrm{k}+3,\mathrm{k}-4,6\mathrm{k}-5\right)$
$\Rightarrow \left(-2+3,2-4,6\left(2\right)-5\right)$
$\Rightarrow \left(1,-2,7\right)$

$=\sqrt{{\left(3-1\right)}^2+{\left(4+2\right)}^2+{\left(4-7\right)}^2}$
$=\sqrt{4+36+9}$
$=\sqrt{49}$