Applying Kirchhoff's Laws to the Electrical Network in the Figure

Applying Kirchhoff's Laws to the electrical network in the figure, the currents I₁, I₂, and I₃ are the solution of the system $$\left\{ \begin{array} { l l } I _ { 1 } - I _ { 2 } + I _ { 3 } & = 0 \\6 I _ { 1 } + 3 I _ { 2 } & = 11 \\3 I _ { 2 } + 5 I _ { 3 } & = 6\end{array} \right.$$ Find the currents. $$\begin{array} { l } V _ { 1 } = 11 \text { volts, } V _ { 2 } = 6 \text { volts } \\R _ { 1 } = 6 \Omega , R _ { 2 } = 3 \Omega , R _ { 3 } = 5 \Omega\end{array}$$

A) I₁ = $- \frac { 40 } { 3 }$ amperes; I₂ = $\frac { 91 } { 3 }$ amperes; I₃ = -17 amperes
B) I₁ = $\frac { 10 } { 9 }$ amperes; I₂ = $\frac { 13 } { 9 }$ amperes; I₃ = $\frac { 1 } { 3 }$ amperes
C) I₁ = 6 amperes; I₂ = 3 amperes; I₃ = 5 amperes
D) I₁ = $\frac { 70 } { 33 }$ amperes; I₂ = $- \frac { 19 } { 33 }$ amperes; I₃ = $\frac { 17 } { 11 }$ amperes
E) I₁ = 70 amperes; I₂ = 91 amperes; I₃ = 21 amperes