Find the coordinates of the point where the line through (3, -4, -5) and (2, -3, 1) crosses the plane determined by the points A(1, 2, 3), B(2, 2, 1) and C(-1, 3, 6).

Equation of line can be found by finding Drs of line we get (3-2,-4+3,-5-1)=(1,-1,-6)Eqn of line can be written in symmetric formx-2/1=y+3/-1=z-1/-6=kGeneral point is (k+2,-k-3,-6k+1)Plane equation is obtained by substituting in ax+by+cz+1=0 all given pointsPlane equation is obtained as 2x+y++z=7..Substitute general point of line in plane to get intersection point..W get k=-1Then substitute in general point We get (1,-2,7)