Let $\frac{x+1}{1}=\frac{y+3}{3}=\frac{-z+2}{2}=k$
Thus, any point on this line will have co-ordinates.
$x=k-1, y=3 k-3, z=-2 k+2$
This line intersects the plane $3 x+4 y+5 z=10$
Since, the value of $x, y, z$ must satisfy equation of plane.
$\begin{array}{ll}\Rightarrow & 3(k-1)+4(3 k-3)+5(-2 k+2)=10 \\ \Rightarrow & 3 k-3+12 k-12+10-10 k=10 \\ \Rightarrow & 5 k-5=10 \\ \Rightarrow & k=3\end{array}$
$\therefore$ The point of intersection is $(x, y, z)$
$\begin{aligned} & \Rightarrow x=3-1, y=3 \times 3-3, z=-2 \times 3+2 \\ & \Rightarrow(x, y, z)=(2,6,-4)\end{aligned}$