If the distance of the point (a, 2, 5) from the image of the point...

Let $P(1, 2, 7)$ be the given point and the given line be $L: \dfrac{x}{1} = \dfrac{y-1}{1} = \dfrac{z-2}{2} = \lambda$.

Any point on the line can be written as $F(\lambda, \lambda+1, 2\lambda+2)$.

If $F$ is the foot of the perpendicular from $P$ to the line, the direction ratios of $PF$ are $(\lambda-1, \lambda-1, 2\lambda-5)$.

Since $PF$ is perpendicular to the line, the dot product of their direction ratios is zero:

$1(\lambda-1) + 1(\lambda-1) + 2(2\lambda-5) = 0$

$6\lambda - 12 = 0 \Rightarrow \lambda = 2$

Thus, the foot of the perpendicular is $F(2, 3, 6)$.

Let the image of $P$ in the line be $P'(x', y', z')$. Since $F$ is the midpoint of $P$ and $P'$:

$\dfrac{x'+1}{2} = 2 \Rightarrow x' = 3$

$\dfrac{y'+2}{2} = 3 \Rightarrow y' = 4$

$\dfrac{z'+7}{2} = 6 \Rightarrow z' = 5$

The image point is $P'(3, 4, 5)$.

The distance of the point $Q(a, 2, 5)$ from $P'(3, 4, 5)$ is given as $4$.

$\sqrt{(a-3)^2 + (2-4)^2 + (5-5)^2} = 4$

$(a-3)^2 + 4 = 16$

$(a-3)^2 = 12$

$a^2 - 6a - 3 = 0$

The sum of all possible values of $a$ is the sum of the roots of this quadratic equation, which is $6$.

Answer: $6$