Given, $${[\matrix{ a & b & c \cr } ]_{1 \times 3}}{\left[ {\matrix{ 1 & 9 & 7 \cr 8 & 2 & 7 \cr 7 & 3 & 7 \cr } } \right]_{3 \times 3}} = [\matrix{ 0 & 0 & 0 \cr } ]$$

$$ \Rightarrow \left[ {\matrix{ {a + 8b + 7c} \cr {9a + 2b + 3c} \cr {7a + 7b + 7c} \cr } } \right] = \left[ {\matrix{ 0 \cr 0 \cr 0 \cr } } \right]$$

$$ \Rightarrow a + 8b + 7c = 0$$ ..... (i)

$$ \Rightarrow 9a + 2b + 3c = 0$$ .... (ii)

$$ \Rightarrow a + b + c = 0$$ .... (iii)

On multiplying Eq. (iii) by 2, then subtract from Eq. (ii), we get

$$7a + c = 0$$ ..... (iv)

Again multiplying Eq. (iii) by 3, then subtract from Eq. (ii), we get

$$6a - b = 0$$ ..... (v)

$$\therefore$$ b = 6a and c = $$-$$ 7a

If b = 6, a = 1 and c = $$-$$7

$$\therefore$$ $$a{x^2} + bx + c = 0$$

$$ \Rightarrow {x^2} + 6x - 7 = 0$$

$$ \Rightarrow (x + 7)(x - 1) = 0$$

$$\therefore$$ $$x = 1,7$$

$$ \Rightarrow \sum\limits_{n = 0}^\infty {{{\left( {{1 \over 1} - {1 \over 7}} \right)}^n} \Rightarrow \sum\limits_{n = 0}^\infty {{{\left( {{6 \over 7}} \right)}^n}} } $$

$$ \Rightarrow 1 + {6 \over 7} + {\left( {{6 \over 7}} \right)^n} + ....\infty $$

$$ = {1 \over {1 - {6 \over 7}}} = {1 \over {1/7}} = 7$$