Let $$(\alpha, \beta, \gamma)$$ be the foot of perpendicular from the point $$(1, 2, 3)$$ on the line $$\frac{x+3}{5} = \frac{y-1}{2} = \frac{z+4}{3}$$. then $$19(\alpha + \beta + \gamma)$$ is equal to :

The line is $$\frac{x+3}{5} = \frac{y-1}{2} = \frac{z+4}{3} = t$$ and a general point on the line is $$(5t-3,\,2t+1,\,3t-4)\,.$$

The foot of the perpendicular from $$(1,2,3)$$ to this line satisfies $$\vec{PF}\cdot\vec{d}=0$$ where $$\vec{d}=(5,2,3)\,.$$

Substituting the coordinates into the dot product gives $$ (5t-3-1)(5) + (2t+1-2)(2) + (3t-4-3)(3) = 0 $$ which simplifies to $$5(5t-4) + 2(2t-1) + 3(3t-7) = 0$$ and to $$25t - 20 + 4t - 2 + 9t - 21 = 0$$ yielding $$38t - 43 = 0 \Rightarrow t = \frac{43}{38}\,.$$

So the foot of the perpendicular $$(\alpha,\beta,\gamma)$$ has coordinates $$\alpha = 5\cdot\frac{43}{38} - 3 = \frac{215 - 114}{38} = \frac{101}{38},\quad \beta = 2\cdot\frac{43}{38} + 1 = \frac{86 + 38}{38} = \frac{124}{38} = \frac{62}{19},\quad \gamma = 3\cdot\frac{43}{38} - 4 = \frac{129 - 152}{38} = \frac{-23}{38}\,.$$

Then $$\alpha+\beta+\gamma = \frac{101}{38} + \frac{124}{38} + \frac{-23}{38} = \frac{202}{38} = \frac{101}{19},$$ and $$19(\alpha+\beta+\gamma) = 19\times\frac{101}{19} = 101\,.$$

The answer is Option (2): $$\boxed{101}$$.