$\textbf{Step 1}$:

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