Let the image of the point P(1, 6, a) in the line L: $$\frac{x}{1} = \frac{y - 1}{2} = \frac{z - a + 1}{b}$$, $$b > 0$$, be $$\left(\frac{a}{3}, 0, a + c\right)$$. If S($$\alpha, \beta, \gamma$$), $$\alpha > 0$$, is the point on L such that the distance of S from the foot of perpendicular from the point P on L is $$2\sqrt{14}$$, then $$\alpha + \beta + \gamma$$ is equal to:

$$L: \frac{x}{1} = \frac{y-1}{2} = \frac{z-a+1}{b}$$

The foot of the perpendicular $$M$$ is the midpoint of the segment joining point $$P(1, 6, a)$$ and its image $$Q\left(\frac{a}{3}, 0, a+c\right)$$:

$$M = \left(\frac{1 + \frac{a}{3}}{2}, \frac{6 + 0}{2}, \frac{a + a + c}{2}\right) = \left(\frac{3+a}{6}, 3, \frac{2a+c}{2}\right)$$

Since $M$ lies on line $$L$$, substituting $$y = 3$$ into the line equation gives: $$\frac{x}{1} = \frac{3-1}{2} = 1 \implies x = 1$$

$$\frac{3+a}{6} = 1 \implies a = 3$$

$$P = (1, 6, 3)$$ and $$M = \left(1, 3, \frac{6+c}{2}\right)$$

$$\frac{z - a + 1}{b} = 1 \implies \frac{\left(\frac{6+c}{2}\right) - 3 + 1}{b} = 1 \implies \frac{c}{2} = b \implies c = 2b$$

$$\vec{PM} = M - P = (1-1, 3-6, 1+b-3) = (0, -3, b-2)$$

The direction vector of line $$L$$ is $$\vec{d} = (1, 2, b)$$. Since vector $$\vec{PM}$$ is perpendicular to line $$L$$: $$\vec{PM} \cdot \vec{d} = 0 \implies (0)(1) + (-3)(2) + (b-2)(b) = 0$$

$$-6 + b^2 - 2b = 0 \implies b^2 - 2b - 6 = 0 \quad \text{--- (1)}$$

Let any general point $$S(\alpha, \beta, \gamma)$$ on line $$L$$ be expressed in terms of parameter $$t$$: $$S = (t, 2t+1, bt + 2)$$

$$SM^2 = (2\sqrt{14})^2 = 56$$

$$(t-1)^2 + (2t+1-3)^2 + (bt+2-(1+b))^2 = 56$$

$$5(t-1)^2 + (b(t-1) + 1)^2 = 56$$

Let $$k = t-1$$: $$5k^2 + b^2k^2 + 2bk + 1 = 56 \implies k^2(5 + b^2) + 2bk - 55 = 0$$

$$k^2(2b + 11) + 2bk - 55 = 0$$

$$2bk(k+1) + 11k^2 - 55 = 0$$

$$b = 3$$  and $$k = 2$$

$$t = 3$$

$$\alpha = t = 3$$

$$\beta = 2t + 1 = 7$$

$$\gamma = bt + 2 = 3(3) + 2 = 11$$

$$\alpha + \beta + \gamma = 3 + 7 + 11 = 21$$