If x = a, y = b, z = c is a solution of the system of linear...

$$\begin{array}{ll}x + 8y + 7z = 0 . . . \left(i\right) \\ 9x + 2y + 3z = 0 . . . \left(ii\right) \\ x + y + z = 0 . . . \left(iii\right)\end{array}$$.....

Solve the equation \left(i\right) \text{ and } \left(iii\right)$</p><p>$⇒ 7y + 6z = 0$</p><p>Solve the equation \left(ii\right) \text{ and } \left(iii\right)$

$⇒ 7x + z = 0$

$x = \lambda \text{ and } \left|y = \dfrac{- 6\left(- 7\lambda\right)}{7}\right| \text{ and } z = - 7\lambda$

$\begin{cases}x = \lambda & \text{ and } y = 6\lambda & \text{ and } z = - 7\lambda\end{cases}$

$$\begin{array}{ll}⇒ \lambda + 12\lambda - 7\lambda = 6 \\ ⇒ 6\lambda = 6 \\ ⇒ \lambda = 1\end{array}$$|$$\begin{array}{ll}⇒ 2\lambda + 6\lambda - 7\lambda \\ ⇒ \lambda = 1\end{array}$$$<br /><br />Hence,$a = 1, b = 6, c = - 7$<br /><br />$∴ 2a + b + c = 1$