$\textbf{Step 1}$:
Given that $x$ = # of members awarded for honesty, $y$ = # of members awarded for for helping others and $z$ = members awarded for supervising the workers to keep the colony neat and clean .
By the given condition; the equations are:
$x+y+z = 12\; (1)$
$2x+3y+3z=33\;(2)$
$x+z=2y \;(3)$
We can express this in the form: $\begin{bmatrix} 1 & 1 & 1\\ 2 & 3 & 3 \\ 1 & -2 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \\ \end {bmatrix}=\begin{bmatrix} 12 \\ 33 \\ 0 \end{bmatrix}$
This is of the form $Ax=B$
Evaluating $A = 1 (3 \times 1 - (-2) \times 3) - 1 (2 \times 1 - 1 \times 3) + 1 (2 \times -2 - 1 \times 3) = =9+1-7=3$
Therefore $A^{-1}$ exists.
$\textbf{Step 2}$:
$A_{11}=3+6=9$
$A_{12}=-2(2-3)=+1$
$A_{13}=-4-3=-7$
$A_{21}=-(1+2)=-3$
$A_{22}=(1-1)=0$
$A_{23}=-(-2-1)=3$
$A_{31}=(3-3)=0$
$A_{32}=-(3-2)=-1$
$A_{33}=(3-2)=1$
Therefore Adjoint of A=$\begin{bmatrix} 9 & -3 & 0\\ 1 & 0 & -1 \\ -7 & 3 & 1 \end{bmatrix}$
Therefore $A^{-1}=\large \frac{1}{3}$$\begin{bmatrix} 9 & -3 & 0\\ 1 & 0 & -1 \\ -7 & 3 & 1 \end{bmatrix}$
$\textbf{Step 3}$:
$AX=B\qquad => X=A^{-1}B$
$\begin{bmatrix} x \\ y \\ z \end {bmatrix} = \frac{1}{3} \begin{bmatrix} 9 & -3 & 0\\ 1 & 0 & -1 \\ -7 & 3 & 1 \end{bmatrix}\begin{bmatrix} 12 \\ 33 \\ 0 \\ \end {bmatrix}=\begin{bmatrix} \frac{9}{3} \\ \frac{12}{3} \\ \frac{15}{3} \\ \end {bmatrix}$
Hence $\begin{bmatrix} x \\ y \\ z \end {bmatrix}=\begin{bmatrix} 3 \\ 4 \\ 5 \end {bmatrix}$
Therefore $x=3,y=4,z=5$
Apart from honesty, cooperaton and supervision, another value which can be added is punctuality among the workers