The point on the line  $\frac{x+2}{2}=\frac{y+6}{3}=\frac{z-34}{-10}$which is nearest to the line $\frac{x+6}{4}=\frac{y-7}{-3}=\frac{z-7}{-2}$is $\left(a,b,c\right)$ where  $a+b+c=$

The given two lines are not parallel lines, so both possess common perpendicular line. The ends of this common perpendicular are the nearest points to each other 

The general point on the line $\frac{x+2}{2}=\frac{y+6}{3}=\frac{z-34}{-10}$ is$P\left(-2+2s,-6+3s,34-10s\right)$

And the general point on the line  $\frac{x+6}{4}=\frac{y-7}{-3}=\frac{z-7}{-2}$is$Q\left(-6+4t,7-3t,7-2t\right)$

The direction ration of line joining  $P,Q$is   $〈-4+4t-2s,13-3t-3s,-27-2t+10s〉$

The line  $PQ$ is perpendicular to both lines 

Hence, 

$\begin{array}{c}2\left(-4+4t-2s\right)+3\left(13-3t-3s\right)-10\left(-27-2t+10s\right)=0\\ 4\left(-4+4t-2s\right)-3\left(13-3t-3s\right)-2\left(-27-2t+10s\right)=0\end{array}$

It gives,$s=3,t=2$

The point  $P\left(4,3,4\right)$is required point. 

Therefore,$a+b+c=4+3+4=\boxed{11}$