If the equation of the plane that contains the point ( - 2, 3, 5 )...

Given

The equation of the plane that contains the point $\left(- 2, 3, 5\right)$ and is perpendicular to each of the planes $2x + 4y + 5z = 8$ and $3x - 2y + 3z = 5$ is $\alphax + \betay + \gammaz + 97 = 0$

Now satisfying the point \left(- 2, 3, 5\right)$in plane$P: \alphax + \betay + \gammaz + 97 = 0$we get,</p><p>$- 2\alpha + 3\beta + 5\gamma + 97 = 0 . . . . . . . . \left(i\right)$</p><p>And using perpendicular condition$2x + 4y + 5z = 8$and$\alphax + \betay + \gammaz + 97 = 0$we get,</p><p>$2\alpha + 4\beta + 5\gamma = 0 . . . . . . \left(ii\right)$</p><p>Similarly,$3\alpha - 2\beta + 3\gamma = 0 . . . . . . . . \left(iii\right)$</p><p>Now solving equation \left(i\right), \left(ii\right) \text{ and } \left(iii\right)$ we get,

$\alpha = 22, \beta = 9 \text{ and } \gamma = - 16$

Hence, $\alpha + \beta + \gamma = 22 + 9 - 16$$= 15$