Question

Let $P_1:x+y+z=0$
$P_2:x+2y+3z=0$
$P_3:x+3y+5z=0$ be three planes
$L_1$ is the line of intersection of $P_1$ & $P_2.\text{Let}P(2,1,-1)$ be point on $P_3$. If $(\alpha ,\beta ,\gamma )$ is image of point $P$ in the line $L_1$ then the value of $|\alpha +\beta +\gamma |$ is

Answer 1

The direction ratio of the line is
$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& 1\\ 1& 2& 3\end{array}\right|=\hat{i}-2\hat{j}+\hat{k}$

Line $L_1$ is $\frac{x}{1}=\frac{y}{-2}=\frac{z}{1}$

Now $(\lambda -2)-2(-2\lambda -1)+(\lambda +1)=0\lambda =-\frac{1}{6}\frac{\alpha +2}{2}=\lambda ,\frac{\beta +1}{2}=-2\lambda ,\frac{\gamma -1}{2}=\lambda \Rightarrow \alpha +\beta +\gamma =-2$