The given system of equations is:

$x+y+z=1$

$2 x+3 y+2 z=2$

$a x+a y+2 a z=4$

The system of equation can be written in the form of $A X=B$, where

$A=\left[\begin{array}{lll}1 & 1 & 1 \\ 2 & 3 & 2 \\ a & a & 2 a\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B=\left[\begin{array}{l}1 \\ 2 \\ 4\end{array}\right]$

Now,

$|A|=1(6 a-2 a)-1(4 a-2 a)+1(2 a-3 a)$

$=4 a-2 a-a=4 a-3 a=a \neq 0$

$\therefore A$ is a non-singular matrix. Therefore, $A^{-1}$ exists.

Hence, the given system of equation is consistent.