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(C) Let the required plane be $P_1$. It contains the line $L_1: \frac{x}{2}=\frac{y}{3}=\frac{z}{4}$,so its normal vector $\vec{n}_1$ is perpendicular

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$8x-y-10z=0$

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(C) Let the required plane be $P_1$. It contains the line $L_1: \frac{x}{2}=\frac{y}{3}=\frac{z}{4}$,so its normal vector $\vec{n}_1$ is perpendicular to $\vec{v}_1 = (2, 3, 4)$.Let $P_2$ be the plane containing the lines $L_2: \frac{x}{3}=\frac{y}{4}=\frac{z}{2}$ and $L_3: \frac{x}{4}=\frac{y}{2}=\frac{z}{3}$.The normal vector $\vec{n}_2$ of $P_2$ is given by the cross product of the direction vectors of $L_2$ and $L_3$:$\vec{n}_2 = (3, 4, 2) 	imes (4, 2, 3) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 4 & 2 \ 4 & 2 & 3 \end{vmatrix} = \hat{i}(12-4) - \hat{j}(9-8) + \hat{k}(6-16) = (8, -1, -10)$.Thus,the equation of plane $P_2$ is $8x - y - 10z = 0$.Since $P_1$ is perpendicular to $P_2$,its normal $\vec{n}_1$ is perpendicular to $\vec{n}_2 = (8, -1, -10)$.Also,$\vec{n}_1$ is perpendicular to $\vec{v}_1 = (2, 3, 4)$.Therefore,$\vec{n}_1 = \vec{v}_1 	imes \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & 4 \ 8 & -1 & -10 \end{vmatrix} = \hat{i}(-30+4) - \hat{j}(-20-32) + \hat{k}(-2-24) = (-26, 52, -26)$.Dividing by $-26$,we get the normal vector $(1, -2, 1)$.The equation of the plane is $1(x-0) - 2(y-0) + 1(z-0) = 0$,which is $x - 2y + z = 0$.

The equation of the plane containing the straight line $\frac{x}{2}=\frac{y}{3}=\frac{z}{4}$ and perpendicular to the plane containing the straight lines $\frac{x}{3}=\frac{y}{4}=\frac{z}{2}$ and $\frac{x}{4}=\frac{y}{2}=\frac{z}{3}$ is:

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