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(C) Let the plane $P$ be $a(x-3) + b(y+4) + c(z-7) = 0$. Since it contains the line with direction ratios $(9, -1, -5)$,we have $9a - b - 5c = 0$. The

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$\frac{32}{3}$

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(C) Let the plane $P$ be $a(x-3) + b(y+4) + c(z-7) = 0$. Since it contains the line with direction ratios $(9, -1, -5)$,we have $9a - b - 5c = 0$. The plane $P$ is perpendicular to the plane containing lines with direction vectors $\vec{v_1} = (2, 3, 5)$ and $\vec{v_2} = (3, 7, 8)$. The normal to this second plane is $\vec{n_2} = \vec{v_1} 	imes \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & 5 \ 3 & 7 & 8 \end{vmatrix} = \hat{i}(24-35) - \hat{j}(16-15) + \hat{k}(14-9) = (-11, -1, 5)$. Since $P$ is perpendicular to this plane,its normal $\vec{n_1} = (a, b, c)$ is perpendicular to $\vec{n_2}$,so $-11a - b + 5c = 0$. Solving the system $9a - b - 5c = 0$ and $-11a - b + 5c = 0$,we add them to get $-2a - 2b = 0$,so $a = -b$. Substituting,$9(-b) - b - 5c = 0 \implies -10b = 5c \implies c = -2b$. Taking $b = -1$,we get $a = 1$ and $c = 2$. The normal vector is $(1, -1, 2)$. The equation of plane $P$ is $1(x-3) - 1(y+4) + 2(z-7) = 0$,which simplifies to $x - y + 2z = 21$. The distance $d$ from $(2, -5, 11)$ to $x - y + 2z - 21 = 0$ is $d = \frac{|2 - (-5) + 2(11) - 21|}{\sqrt{1^2 + (-1)^2 + 2^2}} = \frac{|2 + 5 + 22 - 21|}{\sqrt{6}} = \frac{8}{\sqrt{6}}$. Thus,$d^2 = \frac{64}{6} = \frac{32}{3}$.

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