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(C) Let the equation of the required plane be $ax+by+cz=0$. Since it contains the line $\frac{x}{2}=\frac{y}{3}=\frac{z}{4}$,the normal vector $(a, b,

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$x-2y+z=0$

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(C) Let the equation of the required plane be $ax+by+cz=0$. Since it contains the line $\frac{x}{2}=\frac{y}{3}=\frac{z}{4}$,the normal vector $(a, b, c)$ is perpendicular to the direction vector $(2, 3, 4)$.Thus,$2a+3b+4c=0$ ............$(i)$Next,consider 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 $(a', b', c')$ of this plane is the cross product of the direction vectors $(3, 4, 2)$ and $(4, 2, 3)$:$(a', b', c') = (3, 4, 2) 	imes (4, 2, 3) = (12-4, 8-9, 6-16) = (8, -1, -10)$.Since our required plane is perpendicular to this plane,its normal vector $(a, b, c)$ must be perpendicular to the normal vector $(8, -1, -10)$ of the second plane.Thus,$8a-b-10c=0$ ............$(ii)$Solving $(i)$ and $(ii)$ using cross multiplication:$\frac{a}{(3)(-10) - (4)(-1)} = \frac{b}{(4)(8) - (2)(-10)} = \frac{c}{(2)(-1) - (3)(8)}$$\frac{a}{-30+4} = \frac{b}{32+20} = \frac{c}{-2-24}$$\frac{a}{-26} = \frac{b}{52} = \frac{c}{-26}$Dividing by $-26$,we get the ratios $a:b:c = 1:-2:1$.Substituting these into $ax+by+cz=0$,we get $x-2y+z=0$.

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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