Cramer's Method.
Let determinant Δ=
| 1 8 3 |
| 3 -2 1 | = 1(-4+3)-8(6-2)+3(-9+4)=-1-32-15=-48.
| 2 -3 2 |
This determinant is derived from the coefficients of the 3 variables.
Δ1=
| -3 8 3 |
| -5 -2 1 | = -3(-4+3)-8(-10-6)+3(15+12)=3+128+81=212.
| 6 -3 2 |
I1=Δ1/Δ=212/-48=-53/12.
Δ2=
| 1 -3 3 |
| 3 -5 1 | =1(-10-6)+3(6-2)+3(18+10)=-16+12+84=80.
| 2 6 2 |
I2=Δ2/Δ=80/-48=-5/3.
Δ3=
| 1 8 -3 |
| 3 -2 -5 | = 1(-12-15)-8(18+10)-3(-9+4)=-27-224+15=-236.
| 2 -3 6 |
I3=Δ3/Δ=-236/-48=59/12.
SOLUTION
I1=-53/12, I2=-5/3, I3=59/12.
CHECK
I1+8I2+3I3=-3, 3I1-2I2+I3=-5, 2I1-3I2+2I3=6✔️