3. Given $a^{2}+b c=14, b^{2}-2 b c=-6$. Then $3 a^{2}+4 b^{2}-5 b c=$ $\qquad$
3.18 .
$$
\begin{array}{l}
3 a^{2}+4 b^{2}-5 b c=3\left(a^{2}+b c\right)+4\left(b^{2}-2 b c\right) \\
=3 \times 14+4 \times(-6)=18
\end{array}
$$