If the equation of the plane that contains the point $$(-2, 3, 5)$$ and is perpendicular to each of the planes $$2x + 4y + 5z = 8$$ and $$3x - 2y + 3z = 5$$ is $$\alpha x + \beta y + \gamma z + 97 = 0$$ then $$\alpha + \beta + \gamma =$$

We need a plane containing $$(-2, 3, 5)$$ and perpendicular to both $$2x + 4y + 5z = 8$$ and $$3x - 2y + 3z = 5$$.

Find the normal to the required plane

The normal must be perpendicular to both $$(2, 4, 5)$$ and $$(3, -2, 3)$$:

$$\vec{n} = (2, 4, 5) \times (3, -2, 3) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 4 & 5 \\ 3 & -2 & 3 \end{vmatrix}$$

$$= (12 + 10)\hat{i} - (6 - 15)\hat{j} + (-4 - 12)\hat{k} = 22\hat{i} + 9\hat{j} - 16\hat{k}$$

Write the equation of the plane

$$22(x + 2) + 9(y - 3) - 16(z - 5) = 0$$

$$22x + 44 + 9y - 27 - 16z + 80 = 0$$

$$22x + 9y - 16z + 97 = 0$$

So $$\alpha = 22$$, $$\beta = 9$$, $$\gamma = -16$$.

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