Chọn B
\(P=\sqrt{2} \left|\left(z-z_{1} \right)\left(z-z_{2} \right)\right|+\left|z\left(z-z_{1} \right)\right|+\left|z\left(z-z_{2} \right)\right|\)
Ta có \(\frac{\left(z-z_{2} \right)z}{\left(z_{1} -z_{2} \right)z_{1} } +\frac{z\left(z-z_{1} \right)}{z_{2} \left(z_{2} -z_{1} \right)} +\frac{\left(z-z_{1} \right)\left(z-z_{2} \right)}{z_{1} z_{2} } =1 \)
Lấy module 2 vế ta được:
\(\begin{array}{l} {\, \, \, \, \, \, \, \left|\frac{\left(z-z_{2} \right)z}{\left(z_{1} -z_{2} \right)z_{1} } +\frac{z\left(z-z_{1} \right)}{z_{2} \left(z_{2} -z_{1} \right)} +\frac{\left(z-z_{1} \right)\left(z-z_{2} \right)}{z_{1} z_{2} } \right|=1} \\ {\Rightarrow \frac{\left|z-z_{2} \right|\left|z\right|}{\left|z_{1} -z_{2} \right|\left|z_{1} \right|} +\frac{\left|z\right|\left|z-z_{1} \right|}{\left|z_{2} \right|\left|z_{2} -z_{1} \right|} +\frac{\left|z-z_{1} \right|\left|z-z_{2} \right|}{\left|z_{1} \right|\left|z_{2} \right|} \ge 1} \\ {\Leftrightarrow \frac{\left|z-z_{2} \right|\left|z\right|}{36\sqrt{2} } +\frac{\left|z\right|\left|z-z_{1} \right|}{36\sqrt{2} } +\frac{\left|z-z_{1} \right|\left|z-z_{2} \right|}{36} \ge 1} \\ {\Leftrightarrow P\ge 36\sqrt{2} .} \end{array}\)