1. We are given $[a\ b\ c]\begin{bmatrix}1&9&7\\8&2&7\\7&3&7\end{bmatrix}=[0\ 0\ 0].$ This means $[a\ b\ c]M=0,$ so $(a,b,c)$ satisfies the system obtained by equating each component to zero:

$a+8b+7c=0 \quad ...(1)$ $9a+2b+3c=0 \quad ...(2)$ $7a+7b+7c=0 \quad ...(3)$

Given $a=2$, substitute into these equations.

2. From (3): $7(2+b+c)=0$ $2+b+c=0$ $b+c=-2. \quad ...(4)$

From (1): $2+8b+7c=0$ $8b+7c=-2. \quad ...(5)$

Using (4), $b=-2-c$. Put this in (5): $8(-2-c)+7c=-2$ $-16-8c+7c=-2$ $-16-c=-2$ $c=-14.$ Then $b=-2-(-14)=12.$

So, $a=2,\quad b=12,\quad c=-14.$

3. Now evaluate $\frac{3}{\omega^a}+\frac{1}{\omega^b}+\frac{3}{\omega^c}.$ Since $\omega$ is a cube root of unity with $\operatorname{Im}(\omega)>0$, we have $\omega^3=1,$ and powers reduce modulo $3$.

Compute exponents modulo $3$: $a=2 \implies \omega^a=\omega^2,$ $b=12 \equiv 0 \pmod 3 \implies \omega^{12}=1,$ $c=-14 \equiv 1 \pmod 3 \implies \omega^{-14}=\omega.$ Hence $\frac{3}{\omega^2}+\frac{1}{1}+\frac{3}{\omega^{-14}}=\frac{3}{\omega^2}+1+\frac{3}{\omega}.$

Using $\frac{1}{\omega}=\omega^2, \qquad \frac{1}{\omega^2}=\omega,$ this becomes $3\omega+1+3\omega^2.$

4. Use the identity for cube roots of unity: $1+\omega+\omega^2=0 \implies \omega+\omega^2=-1.$ Therefore, $3\omega+1+3\omega^2=1+3(\omega+\omega^2)=1+3(-1)=-2.$

5. Thus the required value is $\boxed{-2}.$ So the correct option is A.