3. Given that $\triangle A B C$ is an equilateral triangle, its height is
4. If the distances from point $P$ to sides $A B, A C$ are $1, 2$ respectively, then the distance from point $P$ to side $B C$ is ( ).
(A) 1
(B) 1 or 3 or 5
(C) 1 or 3
(D) 1 or 3 or 5 or 7
3. D.

As shown in Figure 5, let point $P$ be inside $\triangle ABC$, and $AH$ be the altitude of $\triangle ABC$. The perpendiculars from $P$ to the three sides of $\triangle ABC$ are $PD$, $PE$, and $PF$, respectively. Let the side length of $\triangle ABC$ be $a$.
From the area relationship, we have
$$
\begin{array}{l}
S_{\triangle PAB} + S_{\triangle PBC} + S_{\triangle PAC} = S_{\triangle ABC} \\
\Rightarrow \frac{1}{2} a PF + \frac{1}{2} a PD + \frac{1}{2} a PE = \frac{1}{2} a AH.
\end{array}
$$

Thus, $1 + PD + 2 = 4 \Rightarrow PD = 1$.
Similarly, when point $P$ is not inside $\triangle ABC$, we can see the positions of $P_i (i=1,2, \cdots, 6)$ in Figure 6.
Since the sum of the distances from $P_j (j=2,3, 4)$ to $AB$ and $AC$ is not less than 4, it is not $P$. Let the distances from $P_i (i=1,5,6)$ to $BC$ be $h_i$. Then
$$
\begin{array}{l}
S_{\triangle P_1AC} + S_{\triangle P_1BC} - S_{\triangle P_1AB} = S_{\triangle ABC}, \\
S_{\triangle P_5AB} + S_{\triangle P_5BC} - S_{\triangle P_5AC} = S_{\triangle ABC}, \\
S_{\triangle P_6BC} - S_{\triangle P_6AB} - S_{\triangle P_6AC} = S_{\triangle ABC}.
\end{array}
$$

Solving these, we get $h_1 = 3$, $h_5 = 5$, $h_6 = 7$.
Therefore, the distance from point $P$ to side $BC$ could be $1, 3, 5, 7$.