Question
The image of the point $$P\left( {\alpha ,\,\beta ,\,\gamma } \right)$$ by the plane $$lx + my + nz = 0$$ is the point $$Q\left( {\alpha ',\,\beta ',\,\gamma '} \right).$$ Then :
A.
$${\alpha ^2} + {\beta ^2} + {\gamma ^2} = {l^2} + {m^2} + {n^2}$$
B.
$${\alpha ^2} + {\beta ^2} + {\gamma ^2} = \alpha {'^2} + \beta {'^2} + \gamma {'^2}$$
C.
$$\alpha \alpha ' + \beta \beta ' + \gamma \gamma ' = 0$$
D.
$$l\left( {\alpha - \alpha '} \right) + m\left( {\beta - \beta '} \right) + n\left( {\gamma - \gamma '} \right) = 0$$
Answer :
$${\alpha ^2} + {\beta ^2} + {\gamma ^2} = \alpha {'^2} + \beta {'^2} + \gamma {'^2}$$
Solution :
Given image of the point $$P\left( {\alpha ,\,\beta ,\,\gamma } \right)$$ by the plane $$lx + my + nz = 0$$ is the point $$Q\left( {\alpha ',\,\beta ',\,\gamma '} \right)$$
But $$lx + my + nz = 0$$ passes through origin.
Let $$O$$ be the origin.
Therefore, $$O$$ is equidistant from $$P$$ and $$Q$$
$$\eqalign{
& \Rightarrow OP = OQ \cr
& \Rightarrow O{P^2} = O{Q^2} \cr
& \therefore \,{\alpha ^2} + {\beta ^2} + {\gamma ^2} = {\alpha ^2} + {\beta ^2} + {\gamma ^2} \cr} $$
Here, option B is correct.