Question

For what value of ${}^{\prime }{k}^{\prime }$ the line $\frac{x-3}{2}=\frac{y+k}{-1}=\frac{z+1}{-5}$ lies on the plane $2x-y+z-7=0$

Answer 1

$\begin{array}{c}Let,\frac{x-3}{2}=\frac{y+k}{-1}=\frac{2+1}{-5}=\lambda \\ \Rightarrow x=2\lambda +3\\ \Rightarrow y=-\lambda -k\\ \Rightarrow z=-5\lambda -1\end{array}$

Putting value of $x,y$ and $z$ in equation $\left(2x-y+z-7=0\right)$

$\begin{array}{c}\Rightarrow 2\left(2\lambda +3\right)-1\left(-\lambda -k\right)+1\left(-5\lambda -1\right)-7=0\\ \Rightarrow 4\lambda +6+\lambda +k-5\lambda -1-7=0\\ \Rightarrow 5\lambda -5\lambda +k-2=0\\ \therefore k=2.\end{array}$

Hence, this is the answer.