The line $\frac{x+3}{3}=\frac{y-2}{-2}=\frac{z+1}{1}$ and the plane $4x+5y+3z-5=0$ intersect at a point 

The given line is $\frac{x+3}{3}=\frac{y-2}{-2}=\frac{z+1}{1}$

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

To get the point of intersection of a line and plane, find value of $k$ such that the above point lies on the plane $4x+5y+3z-5=0$

$\begin{array}{rcl}4\left(3k-3\right)+5\left(-2k+2\right)+3\left(k-1\right)-5& =& 0\\ 12k-12-10k+10+3k-3-5& =& 0\\ 5k-10& =& 0\\ k& =& 2\end{array}$

Therefore, the point of intersection of given line and the plane is $\left(6-3,-4+2,2-1\right)=\boxed{\left(3,-2,1\right)}$