If the length of the perpendicular drawn from the point $$P(a, 4, 2)$$, $$a > 0$$ on the line $$\frac{x+1}{2} = \frac{y-3}{3} = \frac{z-1}{-1}$$ is $$2\sqrt{6}$$ units and Q$$(\alpha_1, \alpha_2, \alpha_3)$$ is the image of the point P in this line, then $$a + \sum_{i=1}^{3} \alpha_i$$ is equal to

We are given a point $$ P(a, 4, 2) $$ with $$ a > 0 $$ and a line given by the equation $$ \frac{x+1}{2} = \frac{y-3}{3} = \frac{z-1}{-1} $$. The length of the perpendicular from P to this line is $$ 2\sqrt{6} $$ units. We need to find the image Q$$ (\alpha_1, \alpha_2, \alpha_3) $$ of P in this line and then compute $$ a + \sum_{i=1}^{3} \alpha_i $$.

First, we express the line in parametric form. Let $$ \frac{x+1}{2} = \frac{y-3}{3} = \frac{z-1}{-1} = \lambda $$. Then the parametric equations are: $$x = 2\lambda - 1$$, $$\quad y = 3\lambda + 3$$, $$\quad z = -\lambda + 1.$$

Let M be the foot of the perpendicular from P to the line. So, M has coordinates $$(2\lambda - 1$$, $$3\lambda + 3$$, $$-\lambda + 1)$$ for some $$ \lambda $$. The direction ratios of the line are (2, 3, -1). Since PM is perpendicular to the line, the vector PM must be perpendicular to the direction vector of the line. The vector PM is: $$ ( (2\lambda - 1) - a, (3\lambda + 3) - 4, (-\lambda + 1) - 2 ) = (2\lambda - 1 - a, 3\lambda - 1, -\lambda - 1). $$

The dot product of PM and the direction vector (2, 3, -1) is zero: $$ (2\lambda - 1 - a) \cdot 2 + (3\lambda - 1) \cdot 3 + (-\lambda - 1) \cdot (-1) = 0. $$

Expanding this: $$ 2(2\lambda - 1 - a) = 4\lambda - 2 - 2a, $$ $$ 3(3\lambda - 1) = 9\lambda - 3, $$ $$ (-1)(-\lambda - 1) = \lambda + 1. $$

Adding them together: $$ 4\lambda - 2 - 2a + 9\lambda - 3 + \lambda + 1 = 0 $$ $$ \Rightarrow 14\lambda - 4 - 2a = 0 $$ $$ \Rightarrow 14\lambda = 2a + 4 $$ $$ \Rightarrow 7\lambda = a + 2 $$ $$ \Rightarrow \lambda = \frac{a + 2}{7}. \quad (1) $$

The distance PM is given as $$ 2\sqrt{6} $$. The distance formula between P(a, 4, 2) and M(2λ - 1, 3λ + 3, -λ + 1) is: $$ \sqrt{ (2\lambda - 1 - a)^2 + (3\lambda + 3 - 4)^2 + (-\lambda + 1 - 2)^2 } = 2\sqrt{6}. $$

Squaring both sides: $$ (2\lambda - 1 - a)^2 + (3\lambda - 1)^2 + (-\lambda - 1)^2 = (2\sqrt{6})^2 = 24. $$

Substituting $$ \lambda = \frac{a + 2}{7} $$ from equation (1): $$ 2\lambda - 1 - a = 2\left(\frac{a + 2}{7}\right) - 1 - a = \frac{2a + 4}{7} - \frac{7}{7} - \frac{7a}{7} = \frac{2a + 4 - 7 - 7a}{7} = \frac{-5a - 3}{7}, $$ $$ 3\lambda - 1 = 3\left(\frac{a + 2}{7}\right) - 1 = \frac{3a + 6}{7} - \frac{7}{7} = \frac{3a - 1}{7}, $$ $$ -\lambda - 1 = -\left(\frac{a + 2}{7}\right) - 1 = \frac{-a - 2}{7} - \frac{7}{7} = \frac{-a - 9}{7}. $$

Substituting these into the distance equation: $$ \left( \frac{-5a - 3}{7} \right)^2 + \left( \frac{3a - 1}{7} \right)^2 + \left( \frac{-a - 9}{7} \right)^2 = 24 $$ $$ \Rightarrow \frac{(-5a - 3)^2 + (3a - 1)^2 + (-a - 9)^2}{49} = 24. $$

Multiplying both sides by 49: $$ (-5a - 3)^2 + (3a - 1)^2 + (-a - 9)^2 = 24 \times 49. $$

Expanding the squares: $$ (-5a - 3)^2 = 25a^2 + 30a + 9, $$ $$ (3a - 1)^2 = 9a^2 - 6a + 1, $$ $$ (-a - 9)^2 = a^2 + 18a + 81. $$

Adding them: $$ 25a^2 + 30a + 9 + 9a^2 - 6a + 1 + a^2 + 18a + 81 = 35a^2 + 42a + 91. $$

Computing $$ 24 \times 49 $$: $$ 24 \times 50 = 1200, \quad 24 \times 1 = 24, \quad \text{so} \quad 1200 - 24 = 1176. $$

Thus: $$ 35a^2 + 42a + 91 = 1176 $$ $$ \Rightarrow 35a^2 + 42a + 91 - 1176 = 0 $$ $$ \Rightarrow 35a^2 + 42a - 1085 = 0. $$

Dividing the entire equation by 7: $$ 5a^2 + 6a - 155 = 0. $$

Solving this quadratic equation using the quadratic formula: $$ a = \frac{ -6 \pm \sqrt{6^2 - 4 \times 5 \times (-155)} }{2 \times 5} = \frac{ -6 \pm \sqrt{36 + 3100} }{10} = \frac{ -6 \pm \sqrt{3136} }{10}. $$

Since $$ \sqrt{3136} = 56 $$ (as $$ 56^2 = 3136 $$): $$ a = \frac{ -6 \pm 56 }{10}. $$

So the solutions are: $$ a = \frac{ -6 + 56 }{10} = \frac{50}{10} = 5, \quad a = \frac{ -6 - 56 }{10} = \frac{-62}{10} = -6.2. $$

Given $$ a > 0 $$, we take $$ a = 5 $$. Thus, point P is (5, 4, 2).

Now, we find the image Q of P in the line. First, find the foot of the perpendicular M using equation (1): $$ \lambda = \frac{a + 2}{7} = \frac{5 + 2}{7} = \frac{7}{7} = 1. $$

So M is: $$x = 2(1) - 1 = 1$$, $$\quad y = 3(1) + 3 = 6$$, $$\quad z = -(1) + 1 = 0.$$ Thus, M is (1, 6, 0).

Since M is the midpoint of P(5, 4, 2) and Q(α₁, α₂, α₃), we have: $$\frac{5 + \alpha_1}{2} = 1$$, $$\quad \frac{4 + \alpha_2}{2} = 6$$, $$\quad \frac{2 + \alpha_3}{2} = 0.$$

Solving each: $$ 5 + \alpha_1 = 2 \Rightarrow \alpha_1 = 2 - 5 = -3, $$ $$ 4 + \alpha_2 = 12 \Rightarrow \alpha_2 = 12 - 4 = 8, $$ $$ 2 + \alpha_3 = 0 \Rightarrow \alpha_3 = -2. $$

So Q is (-3, 8, -2).

Now compute $$ a + \sum_{i=1}^{3} \alpha_i = a + \alpha_1 + \alpha_2 + \alpha_3 $$: $$ 5 + (-3) + 8 + (-2) = 5 - 3 + 8 - 2 = (5 + 8) + (-3 - 2) = 13 - 5 = 8. $$

Hence, the correct answer is Option B.