1) Fixed end moments :

AB =>

$M_{ab} = \frac{-wab^2}{l^2} = \frac{-10*2*3^2}{5^2}$ = -7.2 KN.m

$M_{ba} = \frac{wba^2}{l^2} = \frac{10*3*2^2}{5^2}$ = 4.8 KN.m

BC =>

$M_{bc} = \frac{-wl}{8} = \frac{-12*6}{8}$ = -9 KN.m

$M_{cb} = \frac{wl}{8} = \frac{12*6}{8}$ = 9 KN.m

2) Stiffness (K) :

$K_{BA} = \frac{HEI}{l} = \frac{4EI}{5}$ [$K_{B}$]

$K_{BC} = \frac{HEI(2)}{l} = \frac{4E.2I}{6} = \frac{4EI}{3}$ [$K_{B}$]

Total Stiffness :

$K_{B} = K_{BA} + K_{BC}$

=$\frac{4EI}{5} + \frac{4EI}{5} = \frac{32EI}{15}$

3) Distribution Factor :

$(D.F)_{BA} = \frac{K_{BA}}{K_{B}} = \frac{\frac{4EI}{5}}{\frac{32EI}{15}} = \frac{3}{8}$ [B]

$(D.F)_{BC} = \frac{K_{BC}}{K_{B}} = \frac{\frac{4EI}{3}}{\frac{32EI}{15}} = \frac{5}{8}$ [B]

4) Distribution Factor :