We can write the system of equations in matrix form as:

\begin{bmatrix} 1 & -2 & 4 \ -3 & 1 & -2 \ 2 & -3 & 2 \end{bmatrix} \begin{bmatrix} a \ b \ c \end{bmatrix} = \begin{bmatrix} 8 \ 1 \ -5 \end{bmatrix}

To solve this system using Gauss-Jordan method, we perform row operations on the augmented matrix:

\begin{bmatrix}[ccc

c] 1 & -2 & 4 & 8 \ -3 & 1 & -2 & 1 \ 2 & -3 & 2 & -5 \end{bmatrix}

First, we add 3 times the first row to the second row:

\begin{bmatrix}[ccc

c] 1 & -2 & 4 & 8 \ 0 & -5 & 10 & 25 \ 2 & -3 & 2 & -5 \end{bmatrix}

Next, we subtract 2 times the first row from the third row:

\begin{bmatrix}[ccc

c] 1 & -2 & 4 & 8 \ 0 & -5 & 10 & 25 \ 0 & 1 & -6 & -21 \end{bmatrix}

Then, we multiply the second row by -1/5 to get a leading 1:

\begin{bmatrix}[ccc

c] 1 & -2 & 4 & 8 \ 0 & 1 & -2 & -5 \ 0 & 1 & -6 & -21 \end{bmatrix}

Next, we add the second row to the third row:

\begin{bmatrix}[ccc

c] 1 & -2 & 4 & 8 \ 0 & 1 & -2 & -5 \ 0 & 0 & -8 & -26 \end{bmatrix}

Then, we multiply the third row by -1/8 to get a leading 1:

\begin{bmatrix}[ccc

c] 1 & -2 & 4 & 8 \ 0 & 1 & -2 & -5 \ 0 & 0 & 1 & \frac{13}{4} \end{bmatrix}

Finally, we add 2 times the third row to the second row:

\begin{bmatrix}[ccc

c] 1 & -2 & 4 & 8 \ 0 & 1 & 0 & \frac{1}{2} \ 0 & 0 & 1 & \frac{13}{4} \end{bmatrix}

And we add -4 times the third row to the first row:

\begin{bmatrix}[ccc

c] 1 & -2 & 0 & -2 \ 0 & 1 & 0 & \frac{1}{2} \ 0 & 0 & 1 & \frac{13}{4} \end{bmatrix}

So the solution is:

a = -2 + 2b b = 1/2 c = 13/4

Therefore, the solution to the system of equations is:

a = -2 + 2(1/2) = -1 b = 1/2 c = 13/4