Question

Let $P$be a plane containing the line $\frac{[x-1]}{3}=\frac{[y+6]}{4}=\frac{[z+5]}{2}$ and parallel to the line $\frac{[x-3]}{4}=\frac{[y-2]}{-3}=\frac{[z+5]}{7}$ If the point $(1,-1,\alpha )$lies on the plane $P,$ then the value of $\left|5\alpha \right|$ is equal to

Answer 1

Finding the value of $\left|5\alpha \right|$

Given: $P$be a plane containing the line $\frac{[x-1]}{3}=\frac{[y+6]}{4}=\frac{[z+5]}{2}$ and parallel to the line $\frac{[x-3]}{4}=\frac{[y-2]}{-3}=\frac{[z+5]}{7}$

Equation of the plane is $\left[\begin{array}{ccc}x-1& y+6& z+5\\ 3& 4& 2\\ 4& -3& 7\end{array}\right]$

Also given that $(1,-1,\alpha )$lies on the plane $P,$

$$\begin{gathered} \left[\begin{array}{ccc}1-1& -1+6& \alpha +5\\ 3& 4& 2\\ 4& -3& 7\end{array}\right]=0 \\ \left[\begin{array}{ccc}0& 5& \alpha +5\\ 3& 4& 2\\ 4& -3& 7\end{array}\right]=0 \end{gathered}$$

On Solving the determinant, we get

$$\begin{gathered} 0-5(21-8)+(\alpha +5)(-9-16)=0 \\ -5\times 13+(\alpha +5)(-25)=0 \\ -65-25\alpha -125=0 \\ -190-25\alpha =0 \\ \left|25\alpha \right|=190 \\ \left|5\alpha \right|=\frac{190}{5} \\ \Rightarrow \left|5\alpha \right|=38 \end{gathered}$$

Hence the correct answer is $38$.