1. Write the line in parametric form

Given line: $\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}=t$

So a general point on the line is $x=-3+5t,\quad y=1+2t,\quad z=-4+3t$

Hence the foot of the perpendicular from $(1,2,3)$ on the line is of the form $Q(\alpha,\beta,\gamma)=(-3+5t,\ 1+2t,\ -4+3t)$ for some $t$.

2. Use the perpendicularity condition

A point on the line is $A(-3,1,-4)$ and the direction vector of the line is $\vec d=(5,2,3).$

Given point: $P(1,2,3)$

For $Q$ to be the foot of the perpendicular, vector $\overrightarrow{PQ}$ must be perpendicular to the line's direction vector.

Now, $Q=A+t\vec d$ so $Q=(-3+5t,1+2t,-4+3t).$

Then $\overrightarrow{PQ}=Q-P=(-3+5t-1,\ 1+2t-2,\ -4+3t-3)=(5t-4,\ 2t-1,\ 3t-7).$

Perpendicularity gives $\overrightarrow{PQ}\cdot \vec d=0$

So, $(5t-4)(5)+(2t-1)(2)+(3t-7)(3)=0$

$25t-20+4t-2+9t-21=0$

$38t-43=0$

$t=\frac{43}{38}$

3. Find $\alpha,\beta,\gamma$

$\alpha=-3+5\cdot \frac{43}{38}=-\frac{114}{38}+\frac{215}{38}=\frac{101}{38}$

$\beta=1+2\cdot \frac{43}{38}=\frac{38}{38}+\frac{86}{38}=\frac{124}{38}=\frac{62}{19}$

$\gamma=-4+3\cdot \frac{43}{38}=-\frac{152}{38}+\frac{129}{38}=-\frac{23}{38}$

4. Compute $\alpha+\beta+\gamma$

Using denominator $38$: $\beta=\frac{62}{19}=\frac{124}{38}$

Thus, $\alpha+\beta+\gamma=\frac{101}{38}+\frac{124}{38}-\frac{23}{38}=\frac{202}{38}=\frac{101}{19}$

Therefore, $19(\alpha+\beta+\gamma)=19\cdot \frac{101}{19}=101$

5. Match with options

$19(\alpha+\beta+\gamma)=101$

So the correct option is C.