A line through $(1,-1,2)$ with direction ratios $ \lt 3,2,2 \gt $ meets the plane $x+2 y+3 z=18$. What is the point of intersection of line and plane?

• A

  $(4,4,1)$
• B

  $(2,4,1)$
• Correct answer
  C

  $(4,1,4)$
• D

  $(3,4,7)$

Solution

Let the line $p(1,-1,2)$ and meet the plane at point Q. Direction ratios of the line from the point (1, -1, 2) to the given plane is $\langle 3,2,2\rangle$ $\therefore \quad \frac{x-1}{3}=\frac{y+1}{2}=\frac{z-2}{2}=\lambda$ Co-ordinates any points on the line PQ are $\begin{aligned} & x=3 \lambda+1, y=2 \lambda-1, z=2 \lambda+2 \\ & x+2 y+3 z=18 \\ & \therefore \quad 3 \lambda+1+4 \lambda-2+6 \lambda+6=18 \\ & \Rightarrow \quad 13 \lambda+5=18 \\ & \Rightarrow \quad 13 \lambda=13 \\ & \Rightarrow \quad \lambda \end{aligned}$ Hence, the co-ordinates of Q are $(3+1,2-1,2+2)=(4,1,4)$

Hence, the correct answer is option: 'C'.