$ \text{Step 1: Given Data} $

We are given the forces ${P_1} = 9\,{\rm{kN}}$, ${P_2} = 12\,{\rm{kN}}$, and ${P_3} = 6\,{\rm{kN}}$, the distances ${FE} = 3\,{\rm{m}}$, ${AB} = 3\,{\rm{m}}$, ${BC} = 3\,{\rm{m}}$, ${CD} = 3\,{\rm{m}}$, ${FB} = 3\,{\rm{m}}$, and ${EC} = 3\,{\rm{m}}$.

We are asked to determine the force in members ${EF}$, ${BE}$, ${BC}$ and ${BF}$ of the truss and state if these members are in tension or compression.

$ \text{Step 2: Free Body Diagram} $

First draw the free body diagram for the given system.

$ \text{Step 3: Sign Convention} $

The negative value of force represents compression and the positive value of force represents tension.

Find the reaction force ${N_D}$ by taking moment at point ${A}$ equal to zero using the following relation.
\begin{array}{c}
– {P_1}\left( {AB} \right) – {P_3}\left( {BF} \right) – {P_2}\left( {AB + BC} \right) + {N_D}\left( {AB + BC + CD} \right) = 0
\end{array}

Substitute the known values in the above expression.
\begin{array}{c}
– (9\,\text{kN})(3\,\text{m}) – (6\,\text{kN})(3\,\text{m}) – (12\,\text{kN})(6\,\text{m}) + N_D(9\,\text{m}) = 0 \\
-27\,\text{kN}\cdot\text{m} – 18\,\text{kN}\cdot\text{m} – 72\,\text{kN}\cdot\text{m} + 9N_D\,\text{m} = 0 \\
\hline
N_D = 13\,\text{kN}
\end{array}
$ \text{Step 4: Find } N_D $

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

$ \text{Step 5: Section a-a Analysis} $

Find the length of ${BE}$ using the following relation.
\begin{array}{c}
BE = \sqrt {{{\left( {CE} \right)}^2} + {{\left( {BC} \right)}^2}}
\end{array}

Substitute the known values in the above expression.
\begin{array}{c}
BE = \sqrt {{{\left( {3\,{\rm{m}}} \right)}^2} + {{\left( {3\,{\rm{m}}} \right)}^2}} \\
= 3\sqrt 2 \,{\rm{m}}
\end{array}

$ \text{Step 6: Find } BE $

Find the reaction force ${EF}$ by taking moment at point ${B}$ equal to zero using the following relation.
\begin{array}{c}
– {P_2}\left( {BC} \right) + {N_D}\left( {BC + CD} \right) – {F_{EF}}\left( {EC} \right) = 0
\end{array}

Substitute the known values in the above expression.
\begin{array}{c}
– \left( {12\,{\rm{kN}}} \right)\left( {3\,{\rm{m}}} \right) + \left( {13\,{\rm{kN}}} \right)\left( {3\,{\rm{m}} + 3\,{\rm{m}}} \right) – {F_{EF}}\left( {3\,{\rm{m}}} \right) = 0\\
– 36\,{\rm{kN}} \cdot {\rm{m}} + 78\;{\rm{kN}} \cdot {\rm{m}} + {F_{EF}}\left( {3\,{\rm{m}}} \right) = 0\\
{F_{EF}} = – 14\,{\rm{kN}}\\
{F_{EF}} = 14\,{\rm{kN}}\,\left( {\rm{C}} \right)
\end{array}

$ \text{Step 7: Find } F_{EF} $

Find the reaction force ${BC}$ by taking moment at point ${E}$ equal to zero using the following relation.
\begin{array}{c}
{N_D}\left( {CD} \right) – {F_{BC}}\left( {BC} \right) = 0
\end{array}

Substitute the known values in the above expression.
\begin{array}{c}
\left( {13\,{\rm{kN}}} \right)\left( {3\,{\rm{m}}} \right) – {F_{BC}}\left( {3\,{\rm{m}}} \right) = 0\\
{F_{BC}} = 13\,{\rm{kN}}\,\left( {\rm{T}} \right)
\end{array}

$ \text{Step 8: Find } F_{BC} $

Find the force in member ${BE}$ adding all the forces acting along ${y}$-axis equal to zero using the following relation.
\begin{array}{c}
{N_D} – {F_{BE}}\cos \angle BEC – {P_2} = 0
\end{array}

Substitute the known values in the above expression.
\begin{array}{c}
{N_D} – {F_{BE}}\left( {\frac{{EC}}{{BE}}} \right) – {P_2} = 0
\end{array}

Substitute the known values in the above expression.
\begin{array}{c}
13\,{\rm{kN}} – {F_{BE}}\left( {\frac{{3\,{\rm{m}}}}{{3\sqrt 2 \,{\rm{m}}}}} \right) – 12\,{\rm{kN}} = 0\\
{F_{BE}} = \sqrt 2 \,{\rm{kN}}\\
{F_{BE}} = 1.41\,{\rm{kN}}\,\left( {\rm{T}} \right)
\end{array}

$ \text{Step 9: Find } F_{BF} $

Draw the free body diagram at point B.

Find the force in member ${BF}$ adding all the forces acting along ${y}$-axis at point ${B}$ equal to zero using the following relation.
\begin{array}{c}
{F_{BF}} + {F_{BE}}\cos \angle BEC – {P_1} = 0
\end{array}

Substitute the known values in the above expression.
\begin{array}{c}
{F_{BF}} + {F_{BE}}\left( {\frac{{EC}}{{BE}}} \right) – {P_1} = 0
\end{array}

Substitute the known values in the above expression.
\begin{array}{c}
{F_{BF}} + \left( {1.41\,{\rm{kN}}} \right)\left( {\frac{{3\,{\rm{m}}}}{{3\sqrt 2 \,{\rm{m}}}}} \right) – 9\,{\rm{kN}} = 0\\
{F_{BF}} = 8\,{\rm{kN}}\,\left( {\rm{T}} \right)
\end{array}