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 $$c = - 7a$$

As (a, b, c) lies on $$2x + y + z = 1$$

$$ \Rightarrow 2a + b + c = 1$$

$$ \Rightarrow 2a + 6a - 7a = 1$$

$$\Rightarrow$$ a = 1, b = 6 and c = $$-$$7

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