Since, \(\mathbf{A D} \| \mathbf{B C}\) and \(\mathbf{A D}=2 \mathbf{B C}\)
\(\begin{gathered}
\mathbf{E B} \| \mathbf{F A} \text { and } \mathbf{E B}=2 \mathbf{F A} \\
\mathbf{F C} \| \mathbf{A B} \text { and } \mathbf{F C}=2 \mathbf{A B}
\end{gathered}\)
\(\mathbf{A D}+\mathbf{E B}=2(\mathbf{B C}+\mathbf{F A})=2(\mathbf{A O}+\mathbf{F A})\) ...(i)


In \(\Delta \mathrm{AOF}\),
\(\begin{aligned}
\mathbf{F A}+\mathbf{A O}+\mathbf{O F} & =0 \\
\mathbf{F A}+\mathbf{A O} & =-\mathbf{O F} \quad \ldots (ii)
\end{aligned}\)
Put, Eqs. (ii) in (i)
\(\begin{aligned}
& \mathbf{A D}+\mathbf{E B}=2(-\mathbf{O F})=2 \mathbf{F O} \\
& \mathbf{A D}+\mathbf{E B}=2 \mathbf{A B}
\end{aligned}\)
Now consider,
\(\begin{array}{rlrl}
\mathbf{A D}+\mathbf{E B}+\mathbf{F C} & =2 \mathbf{A B}+2 \mathbf{A B} \quad[\therefore \mathbf{F C}=2 \mathbf{A B}] \\
(3 \lambda-8) \mathbf{A B} & =4 \mathbf{A B} \quad {[\therefore \text { given}]} \\
3 \lambda-8 & =4 \\
3 \lambda & =12 & \\
\lambda & =4 &
\end{array}\)
Hence, option (b) is correct.