Solution:

It is known to us that the eq. of a line passing through two points $A\left(x_{1}, y_{1}, z_{1}\right)$ and $B\left(x_{2}, y_{2}, z_{2}\right)$ is given as
$\frac{\mathrm{X}-\mathrm{X}_{1}}{\mathrm{X}_{2}-\mathrm{X}_{1}}=\frac{\mathrm{y}-\mathrm{y}_{1}}{\mathrm{y}_{2}-\mathrm{y}_{1}}=\frac{\mathrm{Z}-\mathrm{Z}_{1}}{\mathrm{Z}_{2}-\mathrm{Z}_{1}}$
Given: the line passes through the points A $(3,-4,-5)$ and $B(2,-3,1)$
Therefore, $x_{1}=3, y_{1}=-4, z_{1}=-5$
And, $x_{2}=2, y_{2}=-3, z_{2}=1$
Then the eq. of line is
$\begin{array}{l}
\frac{x-3}{2-3}=\frac{y-(-4)}{-3-(-4)}=\frac{z-(-5)}{1-(-5)} \\
\frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+5}{6}=k
\end{array}$
Therefore, $x=-k+3|, y=k-4|, z=6 k-5 \ldots(1)$
Now let $(x, y, z)$ be the coordinates of the point where the line crosses the given plane $2 x+y+z+7=0$
Substituting the value of $x, y, z$ in eq. (1) in the eq. of plane, we obtain
$\begin{array}{l}
2 x+y+z+7=0 \\
2(-k+3)+(k-4)+(6 k-5)=7 \\
5 k-3=7 \\
5 k=10 \\
k=2
\end{array}$
Substituting the value of $k$ in $x, y, z$ we obtain,
$\begin{array}{l}
x=-k+3=-2+3=1 \\
y=k-4=2-4=-2 \\
z=6 k-5=12-5=7
\end{array}$
As a result, the coordinates of the required point are $(1,-2,7)$.