Problem 673

$$\begin{array}{l} \text { Solution: Adding (1) and (2) gives } \\ b^{2}+c^{2}=-a^{2}+14 a-13 \text {. } \\ \text { From (1) we get } b^{2} c^{2}=\left(a^{2}-8 a+7\right)^{2} \text {. } \\ \text { (4)(3) Let } b^{2}=\frac{-a^{2}+14 a-13}{2}+t \text {, } \\ c^{2}=\frac{-a^{2}+14 a-13}{2}-t, \\ \end{array}$$

Substituting into (4) and rearranging gives
$$\begin{array}{l} \left(\frac{-a^{2}+14 a-13}{2}\right)^{2}-\left(a^{2}-8 a+7\right)^{2} \\ =t^{2} \geqslant 0, \end{array}$$

which simplifies to $\left(a^{2}-10 a+9\right)(a-1)^{2} \leqslant 0$.
Solving this, we get $1 \leqslant a \leqslant 9$.
Therefore, the range of values for $a$ is $1 \leqslant a \leqslant 9$.