Find the coordinates of the point where the line passing through $(3, -4, -5)$ and $(2, -3, 1)$ intersects the plane passing through the points $(2, 2, 1)$,$(3, 0, 1)$,and $(4, -1, 0)$.

(A) The equation of the plane passing through the points $A(2, 2, 1)$,$B(3, 0, 1)$,and $C(4, -1, 0)$ is given by the determinant form:
$\begin{vmatrix} x-2 & y-2 & z-1 \\ 3-2 & 0-2 & 1-1 \\ 4-2 & -1-2 & 0-1 \end{vmatrix} = 0$
$\begin{vmatrix} x-2 & y-2 & z-1 \\ 1 & -2 & 0 \\ 2 & -3 & -1 \end{vmatrix} = 0$
$(x-2)(2-0) - (y-2)(-1-0) + (z-1)(-3+4) = 0$
$2(x-2) + 1(y-2) + 1(z-1) = 0$
$2x - 4 + y - 2 + z - 1 = 0$
$2x + y + z - 7 = 0 \ldots(1)$
The equation of the line passing through $(3, -4, -5)$ and $(2, -3, 1)$ is:
$\frac{x-3}{2-3} = \frac{y+4}{-3+4} = \frac{z+5}{1+5} = \lambda$
$\frac{x-3}{-1} = \frac{y+4}{1} = \frac{z+5}{6} = \lambda$
Any point on this line is given by $(x, y, z) = (-\lambda+3, \lambda-4, 6\lambda-5)$.
Since this point lies on the plane $(1)$,we substitute these coordinates into the plane equation:
$2(-\lambda+3) + (\lambda-4) + (6\lambda-5) - 7 = 0$
$-2\lambda + 6 + \lambda - 4 + 6\lambda - 5 - 7 = 0$
$5\lambda - 10 = 0 \implies 5\lambda = 10 \implies \lambda = 2$
Substituting $\lambda = 2$ back into the point coordinates:
$x = -2+3 = 1$
$y = 2-4 = -2$
$z = 6(2)-5 = 7$
Thus,the required point is $(1, -2, 7)$.