70. Given $a: b: c=2: 3: 4, a b+b c+c a=13$, then $b c=$
Answer: 6.
Solution: Clearly $a b c \neq 0$, so
$$
\begin{array}{c}
a b: b c: c a=\frac{a b}{a b c}: \frac{b c}{a b c}: \frac{c a}{a b c}=\frac{1}{c}: \frac{1}{a}: \frac{1}{b}=\frac{1}{4}: \frac{1}{2}: \frac{1}{3}, \\
b c=13 \times \frac{\frac{1}{2}}{\frac{1}{4}+\frac{1}{2}+\frac{1}{3}}=6 .
\end{array}
$$

Alternative solution: Let $a, b, c$ be $2 x, 3 x, 4 x$ respectively, then
$$
\begin{array}{l}
a b+b c+c a \\
=6 x^{2}+12 x^{2}+8 x^{2} \\
=26 x^{2} \\
=13 \\
x^{2}=\frac{1}{2},
\end{array}
$$

Solving for
$$
b c=12 x^{2}=6 \text {. }
$$