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(D) The equation of a plane passing through the point $(-2, 3, 5)$ is given by $a(x + 2) + b(y - 3) + c(z - 5) = 0$.Since this plane is perpendicular

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$15$

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(D) The equation of a plane passing through the point $(-2, 3, 5)$ is given by $a(x + 2) + b(y - 3) + c(z - 5) = 0$.Since this plane is perpendicular to the planes $2x + 4y + 5z = 8$ and $3x - 2y + 3z = 5$,its normal vector $(a, b, c)$ must be perpendicular to the normal vectors of the given planes,which are $\vec{n_1} = (2, 4, 5)$ and $\vec{n_2} = (3, -2, 3)$.Thus,the normal vector $(a, b, c)$ is proportional to the cross product $\vec{n_1} 	imes \vec{n_2}$:$(a, b, c) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 4 & 5 \ 3 & -2 & 3 \end{vmatrix} = \hat{i}(12 - (-10)) - \hat{j}(6 - 15) + \hat{k}(-4 - 12) = 22\hat{i} + 9\hat{j} - 16\hat{k}$.So,$a = 22, b = 9, c = -16$.The equation of the plane is $22(x + 2) + 9(y - 3) - 16(z - 5) = 0$.Expanding this,we get $22x + 44 + 9y - 27 - 16z + 80 = 0$,which simplifies to $22x + 9y - 16z + 97 = 0$.Comparing this with $\alpha x + \beta y + \gamma z + 97 = 0$,we find $\alpha = 22, \beta = 9, \gamma = -16$.Therefore,$\alpha + \beta + \gamma = 22 + 9 - 16 = 15$.

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 $\alpha + \beta + \gamma = ...........$.

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