Find the projection of the line $\frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-3}{4}$ on the plane $x+y+z=7$.

Solution

1. Intersection of line and plane: $$\begin{array}{rl}2t+1-t-1+4t+3& =7⟹5t\\ & =4⟹t\\ & =0.8\end{array}$$. Point $P=(2.6,-1.8,6.2)$.
2. Project point $A(1,-1,3)$ on plane: $$\begin{array}{rl}\frac{x-1}{1}& =\frac{y+1}{1}\\ & =\frac{z-3}{1}\\ & =-\frac{1-1+3-7}{3}\\ & =4/3\end{array}$$.
3. Projected point $$\begin{array}{rl}{A}^{\prime }& =(1+4/3,-1+4/3,3+4/3)\\ & =(7/3,1/3,13/3)\end{array}$$.
4. Direction of projection line is $$\begin{array}{rl}P-A^{\prime }& =(2.6-7/3,-1.8-1/3,6.2-13/3)\\ & =(1/3,-1/3,1/3)\propto (1,-1,1)\end{array}$$.
5. Checking options, the direction vector $(1,1,-2)$ is orthogonal to the normal $(1,1,1)$.
6. Hence the answer is (C).