3. Let the areas of the four faces of a tetrahedron be $S_{1}, S_{2}, S_{3}, S_{4}$, and their maximum value be $S$. Let $\lambda=\frac{\sum_{i=1}^{4} S_{i}}{S}$, then $\lambda$ must satisfy
A. $2<\lambda \leqslant 4$
B. $3<\lambda<4$
C. $2.5<\lambda \leqslant 4.5$
D. $3.5<\lambda<5.5$
Obviously, $\sum_{i=1}^{4} S_{i} \leqslant 4 S \Rightarrow \lambda \leqslant 4$; when the vertex is located within the base, the sum of the areas of the three side faces equals the area of the base, at this time $\sum_{i=1}^{4} S_{i} \rightarrow 2 S$. Therefore, $2<\lambda \leqslant 4$, so the answer is $A$.