Question

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

Answer 1

Equation of line is: $\frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+5}{6}=\lambda$

General point $x=-\lambda +3,y=\lambda +4,z=6\lambda -5$

Equation of plane: $\left|\begin{array}{ccc}x-2& y-2& z-1\\ 3-2& 0-2& 1-1\\ 4-2& -1-2& 0-1\end{array}\right|=0$

$\left|\begin{array}{ccc}x-2& y-2& z-1\\ 1& -2& 0\\ 2& -3& -1\end{array}\right|=0$

$=(x-2)(2-0)-(y-2)(-1-0)+(z-1)(-3+4)=0$

$=2x-4+y-2+z-1=0$

$=2x+y+z-7=0$---- (1)

General point of line is a point on (1),

Therefore, $2(-\lambda +3)+(\lambda -4)+(6\lambda -5)-7=0$

$=-2\lambda +6+\lambda -4+6\lambda -5-7=0$

$=5\lambda -10=0$

$=\lambda =2$

Point of intersection of line and plane $(1,-2,7).$