Equation of Plane Containing Line and Perpendicular to Plane

To find the equation of the required plane, let us proceed step-by-step:

Step 1: Determine the normal vector of the second plane.
The second plane contains the two intersecting straight lines:
$L_2: \frac{x}{3} = \frac{y}{4} = \frac{z}{2}$
$L_3: \frac{x}{4} = \frac{y}{2} = \frac{z}{3}$

The direction vectors of lines $L_2$ and $L_3$ are $\vec{d}_1 = 3\hat{i} + 4\hat{j} + 2\hat{k}$ and $\vec{d}_2 = 4\hat{i} + 2\hat{j} + 3\hat{k}$, respectively.

The normal vector to the second plane, denoted by $\vec{n}_2$, is perpendicular to both $\vec{d}_1$ and $\vec{d}_2$. Therefore, it can be found using the cross product:

$$\vec{n}_2 = \vec{d}_1 \times \vec{d}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 2 \\ 4 & 2 & 3 \end{vmatrix}$$

Evaluating the determinant:
$\vec{n}_2 = \hat{i}(4 \cdot 3 - 2 \cdot 2) - \hat{j}(3 \cdot 3 - 2 \cdot 4) + \hat{k}(3 \cdot 2 - 4 \cdot 4)$
$\vec{n}_2 = \hat{i}(12 - 4) - \hat{j}(9 - 8) + \hat{k}(6 - 16)$
$\vec{n}_2 = 8\hat{i} - \hat{j} - 10\hat{k}$

Step 2: Determine the normal vector of the required plane.
Let the required plane be $P_1$ with normal vector $\vec{n}_1$.
$P_1$ contains the line:
$L_1: \frac{x}{2} = \frac{y}{3} = \frac{z}{4}$
which has the direction vector $\vec{d}_0 = 2\hat{i} + 3\hat{j} + 4\hat{k}$. Thus, $\vec{n}_1$ must be perpendicular to $\vec{d}_0$.

Since plane $P_1$ is perpendicular to the second plane, its normal vector $\vec{n}_1$ must also be perpendicular to $\vec{n}_2$.

Hence, $\vec{n}_1$ is parallel to the cross product $\vec{d}_0 \times \vec{n}_2$:

$$\vec{n}_1 = \vec{d}_0 \times \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 8 & -1 & -10 \end{vmatrix}$$

Evaluating the determinant:
$\vec{n}_1 = \hat{i}(3 \cdot (-10) - 4 \cdot (-1)) - \hat{j}(2 \cdot (-10) - 4 \cdot 8) + \hat{k}(2 \cdot (-1) - 3 \cdot 8)$
$\vec{n}_1 = \hat{i}(-30 + 4) - \hat{j}(-20 - 32) + \hat{k}(-2 - 24)$
$\vec{n}_1 = -26\hat{i} + 52\hat{j} - 26\hat{k}$

Dividing by $-26$, we get a simplified direction vector for the normal to the required plane:
$\vec{n} = \hat{i} - 2\hat{j} + \hat{k}$

Step 3: Formulate the equation of the plane.
Since line $L_1$ passes through the origin $(0, 0, 0)$, the required plane also passes through the origin.

The equation of a plane passing through $(0,0,0)$ with normal vector $\vec{n} = \hat{i} - 2\hat{j} + \hat{k}$ is:
$1(x - 0) - 2(y - 0) + 1(z - 0) = 0$
$x - 2y + z = 0$

Thus, the correct option is C.