If the system of equations x + y + az = b 2x + 5y + 2z = 6 x + 2y +...

Given system of equations are

$x + y + az = b . . . . . \left(i\right)$

$2x + 5y + 2z = 6 . . . . . \left(ii\right)$

$x + 2y + 3z = 3 . . . . . \left(iii\right)$

It is given that the system of equations have infinite solutions.

$⇒ ∆ = 0$

⇒ \left|\begin{cases}1 & 1 & a \\ 2 & 5 & 2 \\ 1 & 2 & 3\end{cases}\right| = 0

$⇒ 1\left(15 - 4\right) - 1\left(6 - 2\right) + a\left(4 - 5\right) = 0$

$⇒ a = 7$

Also, we have $⇒ {∆}_{1} = 0$

⇒ \left|\begin{cases}b & 1 & a \\ 6 & 5 & 2 \\ 3 & 2 & 3\end{cases}\right| = 0

$⇒ b\left(15 - 4\right) - 1\left(18 - 6\right) + 7\left(12 - 15\right) = 0$

$⇒ b = 3$

$∴ a = 7, b = 3$

$∴ 2a + 3b$$= 14 + 9 = 23$

Hence, this is the correct option.