Let [(, , )] be the image of the point [P(2,3,5)] in the plane 2 x+y-3 z=6.

Question

Let $$(\alpha, \beta, \gamma)$$ be the image of the point $$\mathrm{P}(2,3,5)$$ in the plane $2 x+y-3 z=6$. Then $$\alpha+\beta+\gamma$$ is equal to :

Accepted Answer

Co-ordinate of image of point $P(2,3,5)$ in the plane $2 x+y-3 z=6$ is

$$
\begin{aligned}
\frac{\alpha-2}{2} & =\frac{\beta-3}{1}=\frac{\gamma-5}{-3}=\frac{-2(2 \times 2+3-3 \times 5-6)}{2^2+1^2+(-3)^2} \\
\frac{\alpha-2}{2} & =\frac{\beta-3}{1}=\frac{\gamma-5}{-3}=2 \\
\frac{\alpha-2}{2} & =2, \frac{\beta-3}{1}=2, \frac{\gamma-5}{-3}=2 \\
\alpha  & =6, \beta=5, \gamma=-1
\end{aligned}
$$

Hence, $\alpha+\beta+\gamma=6+5-1=10$

Solutions · 1