Four. (This question is worth 18 points) Let $\triangle A B C$ be an equilateral triangle with side length 6. Draw a line $l$ through vertex $A$, and let the distances from vertices $B$ and $C$ to $l$ be denoted as $d_{1}, d_{2}$. Find the maximum value of $d_{1} + d_{2}$.

保留源文本的换行和格式，直接输出翻译结果。
Extend $B A$ to $B^{\prime}$, such that $A B^{\prime}=A B$, and connect $B^{\prime} C$. Then a line $l$ passing through vertex $A$ either intersects $B C$ or intersects $B^{\prime} C$.
(1) If $l$ intersects $B C$ at $D$. Then
$$
\begin{array}{l}
\frac{1}{2}\left(d_{1}+d_{2}\right) \cdot A D \\
=S_{\triangle A B D}+S_{\triangle A D C} \\
=S_{\triangle A B C} \\
=\frac{\sqrt{3}}{4} \cdot 36 .
\end{array}
$$
$$
d_{1}+d_{2}=\frac{18 \sqrt{3}}{A D} \leqslant \frac{18 \sqrt{3}}{3 \sqrt{3}}=6 .
$$

The equality holds if and only if $l \perp B C$.
(2) If $l^{\prime}$ intersects $B^{\prime} C$ at $D^{\prime}$, then
$$
\begin{array}{l}
\frac{1}{2}\left(d_{1}+d_{2}\right) \cdot A D^{\prime}=S_{\triangle B D Y}+S_{\triangle A C D} \\
=S_{\triangle B D^{\prime} A^{\prime}}+S_{\triangle A C C^{\prime}}=S_{\triangle A B C}=S_{\triangle A B C} \\
=\frac{\sqrt{3}}{4} \cdot 36, \\
d_{1}+d_{2}=\frac{18 \sqrt{3}}{A D^{\prime}} \leqslant \frac{18 \sqrt{3}}{3}=6 \sqrt{3} .
\end{array}
$$

The equality holds if and only if $l^{\prime} \perp B^{\prime} C$.
In summary, the maximum value of $d_{1}+d_{2}$ is $6 \sqrt{3}$.