1. We are given $[a\; b\; c]\begin{bmatrix}1&9&7\\8&2&7\\7&3&7\end{bmatrix}=[0\;0\;0].$

This gives a system by equating each column component to $0$: $a+8b+7c=0 \quad ...(1)$ $9a+2b+3c=0 \quad ...(2)$ $7a+7b+7c=0 \quad ...(3)$

Since $b=6$, substitute into these equations: $a+48+7c=0 \Rightarrow a+7c=-48 \quad ...(1')$ $9a+12+3c=0 \Rightarrow 3a+c=-4 \quad ...(2')$ $7a+42+7c=0 \Rightarrow a+c=-6 \quad ...(3')$

2. Solve for $a$ and $c$.

From $(3')$, $a+c=-6.$ From $(2')$, $3a+c=-4.$ Subtracting, $2a=2 \Rightarrow a=1.$ Then $1+c=-6 \Rightarrow c=-7.$

So the quadratic is $ax^2+bx+c=0 \Rightarrow x^2+6x-7=0.$

3. Let roots be $\alpha,\beta$. We need $\sum_{n=0}^{\infty}\left(\frac1\alpha+\frac1\beta\right)^n.$

Using Vieta's formulas for $x^2+6x-7=0,$ we have $\alpha+\beta=-6, \qquad \alpha\beta=-7.$ Therefore, $\frac1\alpha+\frac1\beta=\frac{\alpha+\beta}{\alpha\beta}=\frac{-6}{-7}=\frac67.$

4. The series becomes $\sum_{n=0}^{\infty}\left(\frac67\right)^n.$ This is a geometric series with first term $1$ and common ratio $\frac67$, where $\left|\frac67\right|<1$. Hence, $S=\frac{1}{1-\frac67}=\frac{1}{\frac17}=7.$

5. Therefore the correct option is $\boxed{7}.$