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(B) Let the first line be $L_1$ with direction vector $\vec{v_1} = 2\hat{i} + 3\hat{j} + 4\hat{k}$.Let the two lines in the second plane be $L_2$ and

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$x - 2y + z = 0$

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(B) Let the first line be $L_1$ with direction vector $\vec{v_1} = 2\hat{i} + 3\hat{j} + 4\hat{k}$.Let the two lines in the second plane be $L_2$ and $L_3$ with direction vectors $\vec{v_2} = 3\hat{i} + 4\hat{j} + 2\hat{k}$ and $\vec{v_3} = 4\hat{i} + 2\hat{j} + 3\hat{k}$.The normal vector $\vec{n_2}$ to the second plane is $\vec{v_2} 	imes \vec{v_3} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 4 & 2 \ 4 & 2 & 3 \end{vmatrix} = \hat{i}(12-4) - \hat{j}(9-8) + \hat{k}(6-16) = 8\hat{i} - \hat{j} - 10\hat{k}$.The required plane contains $L_1$ and is perpendicular to the second plane,so its normal vector $\vec{n}$ must be perpendicular to both $\vec{v_1}$ and $\vec{n_2}$.Thus,$\vec{n} = \vec{v_1} 	imes \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & 4 \ 8 & -1 & -10 \end{vmatrix} = \hat{i}(-30+4) - \hat{j}(-20-32) + \hat{k}(-2-24) = -26\hat{i} + 52\hat{j} - 26\hat{k}$.Taking the normal vector as $\hat{i} - 2\hat{j} + \hat{k}$,the equation of the plane passing through the origin $(0,0,0)$ is $1(x-0) - 2(y-0) + 1(z-0) = 0$,which simplifies to $x - 2y + z = 0$.

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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