$ \text{Step 1: Given Data } $

The truss is subjected to the following loads:
\begin{array}{c}
P_1 = 9\,{\rm{kN}} \quad \text{(at point A)} \\
P_2 = 12\,{\rm{kN}} \quad \text{(at point B)} \\
P_3 = 6\,{\rm{kN}} \quad \text{(at point F)}
\end{array}
First draw the free body diagram for the given system.

Key dimensions:
\begin{array}{c}
FE = 3\,{\rm{m}} \\
AB = 3\,{\rm{m}} \\
BC = 3\,{\rm{m}} \\
CD = 3\,{\rm{m}} \\
FB = 3\,{\rm{m}} \\
EC = 3\,{\rm{m}}
\end{array}

$ \text{Step 2: Reaction Force at D} $

Moment equilibrium about A:
\begin{array}{c}
\sum M_A = 0 \\
-P_1(AB) – P_3(FB) – P_2(AB+BC) + N_D(AB+BC+CD) = 0 \\
-9(3) – 6(3) – 12(6) + N_D(9) = 0 \\
-27 – 18 – 72 + 9N_D = 0 \\
\hline
N_D = \frac{117}{9} = 13\,{\rm{kN}}
\end{array}

$ \text{Step 3: Length of BE} $

Now draw the free body diagram for the section $a – a$.

Using Pythagorean theorem:
\begin{array}{c}
BE = \sqrt{EC^2 + BC^2} \\
= \sqrt{3^2 + 3^2} \\
= \sqrt{9 + 9} \\
\hline
BE = 3\sqrt{2}\,{\rm{m}} \approx 4.24\,{\rm{m}}
\end{array}

$ \text{Step 4: Force in EF} $

Moment equilibrium about B:
\begin{array}{c}
\sum M_B = 0 \\
-P_2(BC) + N_D(BC+CD) – F_{EF}(EC) = 0 \\
-12(3) + 13(6) – F_{EF}(3) = 0 \\
-36 + 78 = 3F_{EF} \\
\hline
F_{EF} = 14\,{\rm{kN}}\,(\rm{C}) \\
\text{(Negative value indicates compression)}
\end{array}

$ \text{Step 5: Force in BC} $

Moment equilibrium about E:
\begin{array}{c}
\sum M_E = 0 \\
N_D(CD) – F_{BC}(BC) = 0 \\
13(3) – F_{BC}(3) = 0 \\
\hline
F_{BC} = 13\,{\rm{kN}}\,(\rm{T}) \\
\text{(Positive value indicates tension)}
\end{array}

$ \text{Step 6: Force in BE} $

Vertical equilibrium:
\begin{array}{c}
\sum F_y = 0 \\
N_D – F_{BE}\cos\theta – P_2 = 0 \\
\text{Where } \cos\theta = \frac{EC}{BE} = \frac{3}{3\sqrt{2}} = \frac{1}{\sqrt{2}} \\
13 – F_{BE}(\frac{1}{\sqrt{2}}) – 12 = 0 \\
\hline
F_{BE} = \sqrt{2} \approx 1.41\,{\rm{kN}}\,(\rm{T})
\end{array}

$ \text{Step 7: Force in BF} $
Draw the free body diagram at point B.

Vertical equilibrium at B:
\begin{array}{c}
\sum F_y = 0 \\
F_{BF} + F_{BE}\cos\theta – P_1 = 0 \\
F_{BF} + 1.41(\frac{1}{\sqrt{2}}) – 9 = 0 \\
\hline
F_{BF} = 8\,{\rm{kN}}\,(\rm{T})
\end{array}

$ \text{Step 8: Verification} $

\begin{array}{c}
\text{Moment Check: } \sum M_A = 0 \quad \checkmark \\
\text{Force Balance: } \sum F_x = \sum F_y = 0 \quad \checkmark \\
\hline
\text{All equilibrium conditions are satisfied}
\end{array}

$ \text{Step 9: Final Results} $

\begin{array}{|c|c|c|}
\hline
\text{Member} & \text{Force (kN)} & \text{Nature} \\
\hline
EF & 14 & Compression \\
BE & 1.41 & Tension \\
BC & 13 & Tension \\
BF & 8 & Tension \\
\hline
\end{array}