Given $a=7, c=11, \cos \mathrm{A}=\frac{17}{22}, \cos \mathrm{C}=\frac{1}{14}$
Apply, law of tangent
$$
\tan \left(\frac{\mathrm{c}-\mathrm{A}}{2}\right)=\left(\frac{\mathrm{c}-a}{\mathrm{c}+a}\right) \cot \left(\frac{\mathrm{B}}{2}\right)
$$
Now, $\cos \mathrm{A}=\frac{b^2+c^2-a^2}{2 b c}$
$$
\begin{aligned}
& \frac{17}{22}=\frac{b^2+121-49}{2 b \times 11} \\
& b^2-17 b+72=0 \\
& b^2-9 b-8 b+72=0 \\
& b(b-9)-8(b-9)=0 \\
& (b-8)(b-9)=0 \\
& b=8,9
\end{aligned}
$$
Put $b=9$ in the given expression, Take,
$$
\begin{aligned}
& b \tan \frac{\mathrm{B}}{2} \tan \frac{\mathrm{c}-\mathrm{A}}{2}=\mathrm{b} \tan \frac{\mathrm{B}}{2}\left(\frac{11-7}{11+7}\right) \cot \left(\frac{\mathrm{B}}{2}\right) \\
& =9 \tan \left(\frac{\mathrm{B}}{2}\right) \times \frac{4}{18} \times \frac{1}{\tan \left(\frac{\mathrm{B}}{2}\right)} \\
& b \tan \frac{\mathrm{B}}{2} \tan \left(\frac{\mathrm{c}-\mathrm{A}}{2}\right)=2
\end{aligned}
$$
So, correct option is (c).