= Solution
For $B\gg\Delta_o$, the corresponding <asymptotic expansion> is
$$
\sqrt{225B^2+18B\Delta_o+\Delta_o^2}
=15B+\frac35\Delta_o+\frac{8\Delta_o^2}{375B}+\cdots.
$$
Hence
$$
\boxed{\begin{aligned}
E_0&=-\frac35\Delta_o-\frac{4\Delta_o^2}{375B}+\cdots,\\
E_1&=\frac15\Delta_o,\\
E_2&=15B+\frac{4\Delta_o^2}{375B}+\cdots,\\
E_3&=\frac65\Delta_o.
\end{aligned}}
$$
At zero field, $E_0,E_1,E_3$ converge to the $^3F$ atomic term and $E_2$ to $^3P$ at $15B$. The $^3T_{1g}(P)$ branch therefore has zero initial slope; its high-field slope is not zero.
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