The ratio in which the line joining $(2,-4,3)$ and $(-4,5,-6)$ is divided by the plane $3 x+2 y+z-4=0$ is
- 2 : 1
- 4 : 3
- -1 : 4
- 2 : 3
Solution

By section formula, coordinates of $P=\left(\frac{-4 \lambda+2}{\lambda+1}, \frac{5 \lambda-4}{\lambda+1}, \frac{-6 \lambda+3}{\lambda+1}\right)$ which lies on the plane $3\left(\frac{-4 \lambda+2}{\lambda+1}\right)+2\left(\frac{5 \lambda-4}{\lambda+1}\right)+\left(\frac{-6 \lambda+3}{\lambda+1}\right)=4$ $\begin{array}{lc}\Rightarrow & -12 \lambda+6+10 \lambda-8-6 \lambda+3=4 \lambda+4 \\ \Rightarrow & -8 \lambda+1=4 \lambda+4 \\ \Rightarrow & 12 \lambda=-3 \Rightarrow \lambda=-1 / 4 \\ \Rightarrow & \lambda: 1=-1: 4\end{array}$
Asked in: AP EAMCET 2011
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