$x + y + z = 1,2x + 3y + 2z = 2,ax + ay + 2az = 4.$

The system of equations can be written in the form $AX = B$,

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

Now, $|A| = 1(6a - 2a) - 1(4a - 2a) + 1(2a - 3a) = 4a - 3a = a \ne 0$

Two conditions arise: >
I : If$a \ne 0,$ then $|A| \ne 0$, hence the system of equations is consistent and has a unique solution.

II : If $a = 0$, then $|A| = 0$. So, we need to calculate adj A. Cofactors of elements of A are given by

${A_{11}} = {( - 1)^{1 + 1}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&2\\a&{2a}\end{array}} \right| = 4a,$

${A_{12}} = {( - 1)^{1 + 2}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&2\\a&{2a}\end{array}} \right| = 2a,$

${A_{13}} = {( - 1)^{1 + 3}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&3\\a&a\end{array}} \right| = - a$

${A_{21}} = {( - 1)^{2 + 1}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&1\\a&{2a}\end{array}} \right| = - ( - 4 + 4) = 0$

${A_{22}} = {( - 1)^{2 + 2}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&1\\a&{2a}\end{array}} \right| = a$

${A_{23}} = {( - 1)^{2 + 3}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&1\\a&a\end{array}} \right| = 0$

${A_{31}} = {( - 1)^{3 + 1}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&1\\3&2\end{array}} \right| = - 1$

${A_{32}} = {( - 1)^{3 + 2}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&1\\2&2\end{array}} \right| = 0$

${A_{33}} = {( - 1)^{3 + 3}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&1\\2&3\end{array}} \right| = 1$

% $\therefore$ $adj\;A = {\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{4a}&{ - 2a}&{ - a}\\{ - a}&a&0\\{ - 1}&0&1\end{array}} \right]^{\prime}} = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{4a}&{ - a}&{ - 1}\\{ - 2a}&a&0\\{ - a}&0&1\end{array}} \right]$

Now, $(adj\;A) \cdot B = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{4a}&{ - a}&{ - 1}\\{ - 2a}&a&0\\{ - a}&0&1\end{array}} \right]\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1\\2\\4\end{array}} \right]$

$= \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{2a - 4}\\0\\{ - a + 4}\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 4}\\0\\4\end{array}} \right] \ne 0$

Hence, equations are inconsistent with no solution because if $a = 0$,

then the third system of equations is not possible.