1・32 Given $2 x-3 y-z=0, \quad x+3 y-14 z=0, z \neq 0$, then the value of $\frac{x^{2}+3 x y}{y^{2}+z^{2}}$ is
(A) 7.
(B) 2.
(C) 0.
(D) $-\frac{20}{17}$.
(E) -2.
(15th American High School Mathematics Examination, 1964)
[Solution] From the given, we have $2 x-3 y=z, x+3 y=14 z$,
solving these gives $\quad x=5 z, y=3 z$.
$$
\therefore \quad \frac{x^{2}+3 x y}{y^{2}+z^{2}}=\frac{x(x+3 y)}{y^{2}+z^{2}}=\frac{5 z \cdot 14 z}{9 z^{2}+z^{2}}=\frac{70 z^{2}}{10 z^{2}}=7 \text {. }
$$

Therefore, the answer is (A).