1. Key Concepts and Formulas

• Line of Shortest Distance (LSD): The direction vector of the line of shortest distance between two skew lines with direction vectors $\vec{d_1}$ and $\vec{d_2}$ is given by their cross product, $\vec{d_{LSD}} = \vec{d_1} \times \vec{d_2}$. This vector is perpendicular to both given lines.
• Angle between a Line and a Plane: The angle $\theta$ between a line with direction vector $\vec{d}$ and a plane with normal vector $\vec{n}$ is given by $\sin \theta = \frac{|\vec{d} \cdot \vec{n}|}{|\vec{d}| |\vec{n}|}$. However, in problems where this definition leads to no real solution, it's common for "angle with the plane" to implicitly refer to the angle $\phi$ between the line and the normal of the plane, for which $\cos \phi = \frac{|\vec{d} \cdot \vec{n}|}{|\vec{d}| |\vec{n}|}$. We must choose the interpretation that yields a valid result.
• Image of a Point in a Plane: The image $Q'(\alpha, \beta, \gamma)$ of a point $Q(x_1, y_1, z_1)$ in a plane $Ax+By+Cz+D=0$ can be found using the formula: $\frac{\alpha-x_1}{A} = \frac{\beta-y_1}{B} = \frac{\gamma-z_1}{C} = -2 \frac{Ax_1+By_1+Cz_1+D}{A^2+B^2+C^2}$

2. Step-by-Step Solution

Step 1: Find the Direction Vector of the Line of Shortest Distance (LSD)

• What we're doing: Determining the direction of the line of shortest distance between the two given skew lines.
• Why: The LSD is perpendicular to both skew lines, so its direction vector is the cross product of their direction vectors.
• Given Lines:
  • Line 1 ($L_1$): $\frac{x-2}{0}=\frac{y-1}{1}=\frac{z}{1}$. Its direction vector is $\vec{d_1} = (0, 1, 1)$.
  • Line 2 ($L_2$): $\frac{x-3}{2}=\frac{y-5}{2}=\frac{z-1}{1}$. Its direction vector is $\vec{d_2} = (2, 2, 1)$.
• Calculation: $\vec{d_{LSD}} = \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 1 & 1 \\ 2 & 2 & 1 \end{vmatrix}$ $= \mathbf{i}(1 \cdot 1 - 1 \cdot 2) - \mathbf{j}(0 \cdot 1 - 1 \cdot 2) + \mathbf{k}(0 \cdot 2 - 1 \cdot 2)$ $= \mathbf{i}(1 - 2) - \mathbf{j}(0 - 2) + \mathbf{k}(0 - 2)$ $= -1\mathbf{i} + 2\mathbf{j} - 2\mathbf{k} = (-1, 2, -2)$
• Magnitude of $\vec{d_{LSD}}$: $|\vec{d_{LSD}}| = \sqrt{(-1)^2 + (2)^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$
• Reasoning: The cross product ensures the resulting vector is orthogonal to both input vectors, correctly representing the direction of the LSD.

Step 2: Determine the Value of 'a' for Plane P

• What we're doing: Using the given angle between the LSD and plane P to find the unknown coefficient 'a' in the plane's equation.
• Why: The angle condition provides an equation involving 'a', which can be solved.
• Given Plane: $P: ax - y - z = 0$.
  • The normal vector of plane P is $\vec{n} = (a, -1, -1)$.
  • Magnitude of $\vec{n}$: $|\vec{n}| = \sqrt{a^2 + (-1)^2 + (-1)^2} = \sqrt{a^2+2}$.
• Dot Product: $\vec{d_{LSD}} \cdot \vec{n} = (-1)(a) + (2)(-1) + (-2)(-1) = -a - 2 + 2 = -a$.
• Angle Interpretation: The problem states the angle is $\cos^{-1}\left(\sqrt{\frac{2}{27}}\right)$ with the plane. Let this angle be $\theta$.
  • If $\theta$ is the standard angle between the line and the plane, then $\sin \theta = \frac{|\vec{d_{LSD}} \cdot \vec{n}|}{|\vec{d_{LSD}}| |\vec{n}|}$. Here, $\cos \theta = \sqrt{\frac{2}{27}}$, so $\sin \theta = \sqrt{1 - \left(\sqrt{\frac{2}{27}}\right)^2} = \sqrt{1 - \frac{2}{27}} = \sqrt{\frac{25}{27}} = \frac{5}{3\sqrt{3}}$. Equating, $\frac{5}{3\sqrt{3}} = \frac{|-a|}{3\sqrt{a^2+2}} \implies \frac{5}{\sqrt{3}} = \frac{a}{\sqrt{a^2+2}}$ (since $a>0$, $|-a|=a$). Squaring both sides: $\frac{25}{3} = \frac{a^2}{a^2+2} \implies 25(a^2+2) = 3a^2 \implies 25a^2+50=3a^2 \implies 22a^2=-50 \implies a^2 = -25/11$. This has no real solution for $a$.
  • Therefore, the given angle must refer to the angle $\phi$ between the line and the normal of the plane, for which $\cos \phi = \frac{|\vec{d_{LSD}} \cdot \vec{n}|}{|\vec{d_{LSD}}| |\vec{n}|}$. Given $\phi = \cos^{-1}\left(\sqrt{\frac{2}{27}}\right)$, so $\cos \phi = \sqrt{\frac{2}{27}}$.
• Calculation for 'a': $\sqrt{\frac{2}{27}} = \frac{|-a|}{3\sqrt{a^2+2}}$ Since $a>0$, $|-a|=a$. $\frac{\sqrt{2}}{3\sqrt{3}} = \frac{a}{3\sqrt{a^2+2}}$ $\frac{\sqrt{2}}{\sqrt{3}} = \frac{a}{\sqrt{a^2+2}}$ Square both sides: $\frac{2}{3} = \frac{a^2}{a^2+2}$ $2(a^2+2) = 3a^2$ $2a^2 + 4 = 3a^2$ $a^2 = 4$ Since $a>0$, we get $a=2$.
• Reasoning: Choosing the interpretation that yields a real value for 'a' is a standard approach in such problems when the direct interpretation leads to no solution.

Step 3: Find the Image of the Point in Plane P

• What we're doing: Calculating the coordinates $(\alpha, \beta, \gamma)$ of the image of the given point in the plane P.
• Why: This is a direct application of the image formula, using the value of 'a' found in the previous step.
• Given Point: $Q(1, 1, -5)$.
• Plane P (with $a=2$): $2x - y - z = 0$.
  • Here, $A=2, B=-1, C=-1, D=0$.
• Calculation:
  1. Substitute the coordinates of $Q$ and the plane coefficients into the numerator of the ratio: $Ax_1+By_1+Cz_1+D = 2(1) + (-1)(1) + (-1)(-5) + 0 = 2 - 1 + 5 = 6$.
  2. Substitute the coefficients into the denominator of the ratio: $A^2+B^2+C^2 = 2^2 + (-1)^2 + (-1)^2 = 4 + 1 + 1 = 6$.
  3. Calculate the common ratio ($k$): $k = -2 \frac{Ax_1+By_1+Cz_1+D}{A^2+B^2+C^2} = -2 \frac{6}{6} = -2$
  4. Solve for $\alpha, \beta, \gamma$:
     • $\frac{\alpha-1}{2} = -2 \implies \alpha-1 = -4 \implies \alpha = -3$.
     • $\frac{\beta-1}{-1} = -2 \implies \beta-1 = 2 \implies \beta = 3$.
     • $\frac{\gamma-(-5)}{-1} = -2 \implies \gamma+5 = 2 \implies \gamma = -3$.
• Image Point: $Q'(\alpha, \beta, \gamma) = (-3, 3, -3)$.
• Reasoning: The image formula directly computes the reflected point by ensuring the midpoint lies on the plane and the line connecting the point and its image is normal to the plane.

Step 4: Calculate the Final Expression

• What we're doing: Substituting the found values of $\alpha, \beta, \gamma$ into the required expression.
• Why: This is the final step to obtain the numerical answer.
• Values: $\alpha = -3$, $\beta = 3$, $\gamma = -3$.
• Calculation: $\alpha+\beta-\gamma = (-3) + (3) - (-3)$ $= -3 + 3 + 3$ $= 3$
• Reasoning: This is a straightforward algebraic substitution.

3. Common Mistakes & Tips

• Angle Interpretation: Always check both interpretations of "angle with the plane" (with the plane itself using $\sin \theta$, and with the normal vector using $\cos \phi$) if the initial interpretation leads to no real solutions. The problem setter often intends the one that yields a valid answer.
• Sign Errors: Be meticulous with signs, especially when calculating cross products, dot products, and substituting into the image formula (e.g., the $-2$ factor and handling negative coordinates).
• Formula Recall: Ensure accurate recall of the image of a point in a plane formula, as any error in its application can propagate through the entire calculation.

4. Summary

This problem required a sequential application of several 3D geometry concepts. First, we found the direction vector of the line of shortest distance by taking the cross product of the direction vectors of the two given skew lines. Next, we used the angle condition between this LSD and the plane P to determine the value of 'a'. Given the problem's constraints, the angle was interpreted as that between the LSD and the normal vector of the plane, yielding $a=2$. Finally, we used the formula for the image of a point in a plane to find the coordinates $(\alpha, \beta, \gamma)$ of the image of $(1,1,-5)$ in the plane $2x-y-z=0$, which resulted in $(-3, 3, -3)$. Substituting these values into the expression $\alpha+\beta-\gamma$ gave the final answer of 3.

The final answer is $\boxed{3}$.