3. In $\triangle A B C$, it is known that
$$
\angle A=30^{\circ}, 2 \overrightarrow{A B} \cdot \overrightarrow{A C}=3 \overrightarrow{B C}^{2} \text {. }
$$

Then the cosine of the largest angle of $\triangle A B C$ is
3. $-\frac{1}{2}$.

Let $A B=c, A C=b, B C=a$.
From $2 \overrightarrow{A B} \cdot \overrightarrow{A C}=3 \overrightarrow{B C}^{2} \Rightarrow 2 c b \cos A=3 a^{2}$.
Since $\angle A=30^{\circ}$, we have $b c=\sqrt{3} a^{2}$.
Combining with the sine rule, we get
$$
\begin{array}{l}
\sin B \cdot \sin C=\sqrt{3} \sin ^{2} A=\frac{\sqrt{3}}{4} \\
\Rightarrow \cos (C-B)-\cos (C+B)=\frac{\sqrt{3}}{2} \text {. } \\
\text { Also } \angle C+\angle B=\pi-\angle A=150^{\circ} \text {, then } \\
\cos (C-B)=0 \\
\Rightarrow |\angle C-\angle B|=90^{\circ} \\
\Rightarrow \angle C=120^{\circ}, \angle B=30^{\circ}
\end{array}
$$

or $\angle C=30^{\circ}, \angle B=120^{\circ}$.
Therefore, the cosine value of the largest angle in $\triangle A B C$ is $-\frac{1}{2}$.