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(C) First,find the equation of the plane containing the lines $\frac{x}{3}=\frac{y}{4}=\frac{z}{2}$ and $\frac{x}{4}=\frac{y}{2}=\frac{z}{3}$. The nor

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$x-2y+z=0$

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(C) First,find the equation of the plane containing the lines $\frac{x}{3}=\frac{y}{4}=\frac{z}{2}$ and $\frac{x}{4}=\frac{y}{2}=\frac{z}{3}$. The normal vector $\vec{n_1}$ to this plane is given by the cross product of the direction vectors $\vec{v_1} = (3, 4, 2)$ and $\vec{v_2} = (4, 2, 3)$.$\vec{n_1} = \vec{v_1} 	imes \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 4 & 2 \ 4 & 2 & 3 \end{vmatrix} = \hat{i}(12-4) - \hat{j}(9-8) + \hat{k}(6-16) = 8\hat{i} - \hat{j} - 10\hat{k}$.The equation of this plane is $8x - y - 10z = 0$.Let the required plane be $ax + by + cz = 0$. Since it contains the line $\frac{x}{2}=\frac{y}{3}=\frac{z}{4}$,its normal vector $\vec{n_2} = (a, b, c)$ must be perpendicular to the line's direction vector $\vec{v_3} = (2, 3, 4)$. Thus,$2a + 3b + 4c = 0$.Since the required plane is perpendicular to the plane $8x - y - 10z = 0$,their normal vectors are perpendicular: $(a, b, c) \cdot (8, -1, -10) = 0$,so $8a - b - 10c = 0$.Solving these two equations for $a, b, c$ using cross product: $\vec{n_2} = (3, 4, 2) 	imes (8, -1, -10) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & 4 \ 8 & -1 & -10 \end{vmatrix} = \hat{i}(-30+4) - \hat{j}(-20-32) + \hat{k}(-2-24) = -26\hat{i} + 52\hat{j} - 26\hat{k}$.Dividing by $-26$,we get the normal vector $(1, -2, 1)$.The equation of the plane is $1(x) - 2(y) + 1(z) = 0$,which is $x - 2y + z = 0$.

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