1. Write the two lines in parametric form

For the line $\frac{x+6}{3}=\frac{y}{2}=\frac{z+1}{1}=t,$ we get $x=3t-6,\quad y=2t,\quad z=t-1.$

For the line $\frac{x-7}{4}=\frac{y-9}{3}=\frac{z-4}{2}=s,$ we get $x=4s+7,\quad y=3s+9,\quad z=2s+4.$

2. Find their point of intersection

At the intersection, coordinates must be equal: $3t-6=4s+7 \quad ...(1)$ $2t=3s+9 \quad ...(2)$ $t-1=2s+4 \quad ...(3)$

From (3): $t=2s+5.$

Substitute into (2): $2(2s+5)=3s+9$ $4s+10=3s+9$ $s=-1.$ Then $t=2(-1)+5=3.$

Now substitute into either line to get the intersection point: $x=3(3)-6=3,$ $y=2(3)=6,$ $z=3-1=2.$

So the point of intersection is $(3,6,2).$

3. Find the distance from $(7,8,9)$

Distance squared is $d^2=(3-7)^2+(6-8)^2+(2-9)^2$ $=(-4)^2+(-2)^2+(-7)^2$ $=16+4+49=69.$

Thus, $d^2+6=69+6=75.$

4. Match with the options

The correct option is $\boxed{75}$ which is Option A.