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

Question

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

Accepted Answer

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

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

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

Solutions · 1