The line of shortest distance between the lines $$\dfrac{x-2}{0} = \dfrac{y-1}{1} = \dfrac{z}{1}$$ and $$\dfrac{x-3}{2} = \dfrac{y-5}{2} = \dfrac{z-1}{1}$$ makes an angle of $$\sin^{-1}\sqrt{\dfrac{2}{27}}$$ with the plane $$P: ax - y - z = 0$$, $$a > 0$$. If the image of the point $$(1, 1, -5)$$ in the plane $$P$$ is $$(\alpha, \beta, \gamma)$$, then $$\alpha + \beta - \gamma$$ is equal to ______.

We need to find $$\alpha + \beta - \gamma$$ where $$(\alpha, \beta, \gamma)$$ is the image of $$(1, 1, -5)$$ in plane $$P: ax - y - z = 0$$.

To determine the line of shortest distance between the given lines, we note the following.

The first line is given by $$\dfrac{x-2}{0} = \dfrac{y-1}{1} = \dfrac{z}{1}$$, which passes through $$(2, 1, 0)$$ with direction vector $$\vec{d_1} = (0, 1, 1)$$.

The second line is given by $$\dfrac{x-3}{2} = \dfrac{y-5}{2} = \dfrac{z-1}{1}$$, which passes through $$(3, 5, 1)$$ with direction vector $$\vec{d_2} = (2, 2, 1)$$.

The direction of the shortest distance line is given by the cross product $$\vec{d_1} \times \vec{d_2}$$:

$$\vec{d_1} \times \vec{d_2} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 0 & 1 & 1 \\ 2 & 2 & 1 \end{vmatrix} = (1-2)\vec{i} - (0-2)\vec{j} + (0-2)\vec{k} = (-1, 2, -2)$$

We then determine the angle that this line makes with the plane $$P: ax - y - z = 0$$.

The normal vector to $$P$$ is $$\vec{n} = (a, -1, -1)$$.

The sine of the angle between a line with direction $$\vec{d}$$ and a plane with normal $$\vec{n}$$ is:

$$\sin \theta = \dfrac{|\vec{d} \cdot \vec{n}|}{|\vec{d}||\vec{n}|}$$

Since $$\theta = \sin^{-1}\sqrt{\dfrac{2}{27}}$$, it follows that $$\sin \theta = \sqrt{\dfrac{2}{27}}$$.

$$\vec{d} \cdot \vec{n} = (-1)(a) + (2)(-1) + (-2)(-1) = -a - 2 + 2 = -a$$ $$|\vec{d}| = \sqrt{1 + 4 + 4} = 3, \quad |\vec{n}| = \sqrt{a^2 + 1 + 1} = \sqrt{a^2 + 2}$$

Accordingly,

$$\sqrt{\dfrac{2}{27}} = \dfrac{|{-a}|}{3\sqrt{a^2 + 2}} = \dfrac{a}{3\sqrt{a^2 + 2}}$$ (since $$a > 0$$)

Squaring both sides gives $$\dfrac{2}{27} = \dfrac{a^2}{9(a^2 + 2)}$$

$$2 \cdot 9(a^2 + 2) = 27a^2$$ $$18a^2 + 36 = 27a^2$$ $$9a^2 = 36$$ $$a^2 = 4 \implies a = 2$$ (since $$a > 0$$)

Substituting $$a = 2$$, we now find the image of $$(1, 1, -5)$$ in the plane $$2x - y - z = 0$$.

Recall the reflection formula for a point $$(x_0, y_0, z_0)$$ in the plane $$Ax + By + Cz + D = 0$$:

$$\dfrac{\alpha - x_0}{A} = \dfrac{\beta - y_0}{B} = \dfrac{\gamma - z_0}{C} = \dfrac{-2(Ax_0 + By_0 + Cz_0 + D)}{A^2 + B^2 + C^2}$$

Here, $$A = 2, B = -1, C = -1, D = 0$$ and $$(x_0, y_0, z_0) = (1, 1, -5)$$:

$$Ax_0 + By_0 + Cz_0 + D = 2 - 1 + 5 = 6$$ $$A^2 + B^2 + C^2 = 4 + 1 + 1 = 6$$ $$\dfrac{-2(6)}{6} = -2$$ $$\alpha = 1 + 2(-2) = -3$$ $$\beta = 1 + (-1)(-2) = 3$$ $$\gamma = -5 + (-1)(-2) = -3$$

Finally, summing these values yields:

$$\alpha + \beta - \gamma = -3 + 3 - (-3) = -3 + 3 + 3 = 3$$

The answer is $$3$$.