3. Given that $P$ is a point inside $\triangle A B C$ and satisfies $2 P A+3 P B+4 P C=0$.
Then, $S_{\triangle P B C}: S_{\triangle P C A}: S_{\triangle P A B}$ equals ( ).
(A) $1: 2: 3$
(B) $2: 3: 4$
(C) $3: 4: 2$
(D) $4: 3: 2$
3. B.

As shown in Figure 4, extend $PA$ to $D$ such that $PD=2PA$; extend $PB$ to $E$ such that $PE=3PB$; extend $PC$ to $F$ such that $PF=4PC$. Then,
$$
PD + PE + PF = 0.
$$

Thus, $P$ is the centroid of $\triangle DEF$. Therefore, we have
$$
\begin{array}{l}
S_{\triangle PBC} = \frac{1}{3 \times 4} S_{\triangle PEF} \\
= \frac{1}{3} \times \frac{1}{3 \times 4} S_{\triangle DEF} = \frac{1}{36} S_{\triangle DEF}, \\
S_{\triangle PCA} = \frac{1}{4 \times 2} S_{\triangle PFD} = \frac{1}{3} \times \frac{1}{4 \times 2} S_{\triangle DEF} = \frac{1}{24} S_{\triangle DEF}, \\
S_{\triangle PAB} = \frac{1}{2 \times 3} S_{\triangle PDE} = \frac{1}{3} \times \frac{1}{2 \times 3} S_{\triangle DEF} = \frac{1}{18} S_{\triangle DEF}. \\
\text{Hence, } S_{\triangle PBC} : S_{\triangle PCA} : S_{\triangle PAB} = \frac{1}{36} : \frac{1}{24} : \frac{1}{18} = 2 : 3 : 4.
\end{array}
$$