This problem involves two main steps: first, finding the mirror image of a point with respect to a line, and second, determining the equation of a plane that passes through this mirror image and contains another given line.

1. Key Concepts and Formulas

• Mirror Image of a Point with respect to a Line: If a point $A(x_1, y_1, z_1)$ has a mirror image $B(x_2, y_2, z_2)$ with respect to a line $L$, then the foot of the perpendicular $M$ from $A$ to $L$ is the midpoint of the segment $AB$. Additionally, the line segment $AM$ is perpendicular to the line $L$.
  • To find $M$: Let $M$ be a general point on $L$. Form the vector $\vec{AM}$. Set the dot product of $\vec{AM}$ and the direction vector of $L$ to zero to find the parameter for $M$.
  • To find $B$: Use the midpoint formula: $M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right)$.
• Equation of a Plane containing a Line and a Point: A plane containing a line (which passes through a point $P_L$ and has direction vector $\vec{d_L}$) and also passing through an external point $P_E$ can be found as follows:
  • Two vectors lying in the plane are $\vec{d_L}$ and $\vec{P_L P_E}$ (or $\vec{P_E P_L}$).
  • The normal vector to the plane, $\vec{n}$, is perpendicular to both of these vectors. Thus, $\vec{n} = \vec{d_L} \times \vec{P_L P_E}$.
  • The equation of the plane is $\vec{n} \cdot (\vec{r} - \vec{P_E}) = 0$ or $Ax + By + Cz = D$, where $(A, B, C)$ are the components of $\vec{n}$, and $D$ is found by substituting the coordinates of $P_E$ (or $P_L$) into the equation.

2. Step-by-Step Solution

Step 1: Identify the given points and lines. Let the given point be $A = (2, 3, 1)$. The equation of the line $L_1$ (with respect to which the mirror image is taken) is: ${{x + 1} \over 2} = {{y - 3} \over 1} = {{z + 2} \over { - 1}}$ We can write a general point on $L_1$ as $M = (-1 + 2\lambda, 3 + \lambda, -2 - \lambda)$. The direction vector of $L_1$ is $\vec{d_1} = (2, 1, -1)$.

The equation of the line $L_2$ (contained in the plane) is: ${{x - 2} \over 3} = {{1 - y} \over 2} = {{z + 1} \over 1}$ To get it into standard form, we rewrite $1-y$ as $-(y-1)$: ${{x - 2} \over 3} = {{y - 1} \over { - 2}} = {{z + 1} \over 1}$ A point on $L_2$ is $P_{L2} = (2, 1, -1)$. The direction vector of $L_2$ is $\vec{d_2} = (3, -2, 1)$.

Step 2: Find the mirror image of point A with respect to line $L_1$. Let $B(x_B, y_B, z_B)$ be the mirror image of $A(2, 3, 1)$. Let $M$ be the foot of the perpendicular from $A$ to $L_1$. $M$ is a point on $L_1$, so $M = (-1 + 2\lambda, 3 + \lambda, -2 - \lambda)$. The vector $\vec{AM}$ is perpendicular to the direction vector $\vec{d_1}$ of $L_1$. $\vec{AM} = M - A = (-1 + 2\lambda - 2, 3 + \lambda - 3, -2 - \lambda - 1) = (2\lambda - 3, \lambda, - \lambda - 3)$. Since $\vec{AM} \cdot \vec{d_1} = 0$: $(2\lambda - 3)(2) + (\lambda)(1) + (-\lambda - 3)(-1) = 0$ $4\lambda - 6 + \lambda + \lambda + 3 = 0$ $6\lambda - 3 = 0 \implies \lambda = \frac{1}{2}$ Substitute $\lambda = \frac{1}{2}$ into the coordinates of $M$: $M = \left(-1 + 2\left(\frac{1}{2}\right), 3 + \frac{1}{2}, -2 - \frac{1}{2}\right) = \left(-1 + 1, \frac{7}{2}, -\frac{5}{2}\right) = \left(0, \frac{7}{2}, -\frac{5}{2}\right)$ Since $M$ is the midpoint of $AB$: $\frac{2 + x_B}{2} = 0 \implies 2 + x_B = 0 \implies x_B = -2$ $\frac{3 + y_B}{2} = \frac{7}{2} \implies 3 + y_B = 7 \implies y_B = 4$ $\frac{1 + z_B}{2} = -\frac{5}{2} \implies 1 + z_B = -5 \implies z_B = -6$ So, the mirror image point is $B = (-2, 4, -6)$.

Step 3: Find the equation of the plane passing through B and containing line $L_2$. The plane passes through $B(-2, 4, -6)$ and contains line $L_2$, which passes through $P_{L2}(2, 1, -1)$ and has direction vector $\vec{d_2} = (3, -2, 1)$. A vector lying in the plane is $\vec{P_{L2}B} = B - P_{L2} = (-2 - 2, 4 - 1, -6 - (-1)) = (-4, 3, -5)$. The normal vector $\vec{n}$ to the plane is perpendicular to both $\vec{d_2}$ and $\vec{P_{L2}B}$. We can find $\vec{n}$ using the cross product: $\vec{n} = \vec{d_2} \times \vec{P_{L2}B} = (3, -2, 1) \times (-4, 3, -5)$ $\vec{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & -2 & 1 \\ -4 & 3 & -5 \end{vmatrix}$ $\vec{n} = \mathbf{i}((-2)(-5) - (1)(3)) - \mathbf{j}((3)(-5) - (1)(-4)) + \mathbf{k}((3)(3) - (-2)(-4))$ $\vec{n} = \mathbf{i}(10 - 3) - \mathbf{j}(-15 + 4) + \mathbf{k}(9 - 8)$ $\vec{n} = 7\mathbf{i} - (-11)\mathbf{j} + 1\mathbf{k} = (7, 11, 1)$ The equation of the plane is of the form $Ax + By + Cz = D$, where $(A, B, C)$ are the components of the normal vector. So, the equation is $7x + 11y + z = D$. To find $D$, substitute the coordinates of point $B(-2, 4, -6)$ into the equation: $7(-2) + 11(4) + (-6) = D$ $-14 + 44 - 6 = D$ $24 = D$ Thus, the equation of the plane is $7x + 11y + z = 24$.

Step 4: Determine $\alpha, \beta, \gamma$ and calculate their sum. Comparing the obtained plane equation $7x + 11y + z = 24$ with the given form $\alpha x + \beta y + \gamma z = 24$, we get: $\alpha = 7$ $\beta = 11$ $\gamma = 1$ Now, we calculate the sum $\alpha + \beta + \gamma$: $\alpha + \beta + \gamma = 7 + 11 + 1 = 19$

3. Common Mistakes & Tips

• Sign Errors in Line Equations: Always rewrite line equations into the standard form {{x - x_0} \over l} = {{y - y_0} \over m} = {{z - z_0} \over n}} to correctly identify the point $(x_0, y_0, z_0)$ and the direction vector $(l, m, n)$. For example, $1-y$ must be written as $-(y-1)$, changing the sign of the corresponding direction component.
• Cross Product Calculation: Be meticulous with the signs and order of terms when calculating the cross product, as a single error can propagate through the rest of the problem.
• Perpendicularity Condition: Remember that the vector from the point to the foot of the perpendicular on the line is perpendicular to the line's direction vector. This is a crucial step in finding the mirror image.

4. Summary We first found the mirror image of the given point $(2, 3, 1)$ with respect to the first line by using the conditions of perpendicularity and midpoint. The mirror image point was found to be $(-2, 4, -6)$. Next, we determined the equation of the plane passing through this mirror image point and containing the second given line. We achieved this by finding two vectors in the plane (the direction vector of the line and a vector connecting the mirror image point to a point on the line) and calculating their cross product to get the normal vector of the plane. The equation of the plane was $7x + 11y + z = 24$. Comparing this with the given form $\alpha x + \beta y + \gamma z = 24$, we found $\alpha = 7, \beta = 11, \gamma = 1$. Finally, their sum was calculated as $7 + 11 + 1 = 19$.

The final answer is $\boxed{19}$, which corresponds to option (B).