Solve the system of equations

$$
\frac{1}{x}+\frac{1}{y}=\frac{1}{z} ; \quad \frac{1}{x+15}+\frac{1}{y-6}=\frac{1}{z} ; \quad \frac{1}{x+24}+\frac{1}{y-15}=\frac{1}{z}
$$
Clearly $x \neq 0, -15, -24$ and $y \neq 0, 6, 15$ and $z \neq 0$. Furthermore, from $0 \neq \frac{1}{z}=\frac{x+y+9}{(x+15)(y-6)}$ it follows that $x+y+9 \neq 0$. Let's introduce the new variables $x+15=u$ and $y-6=v$. Then, from the equality of the right-hand sides of equations (2) and (3):

$$
\frac{1}{u}+\frac{1}{v}=\frac{1}{u+9}+\frac{1}{v-9}
$$

Bringing to a common denominator, we get the equation

$$
\frac{u+v}{u v}=\frac{u+v}{(u+9)(v-9)}
$$

Under the condition $u+v=x+y+9 \neq 0$, we can divide both sides of the equation by $u+v$ and take the reciprocals; we get

$$
u v=u v+9 v-9 u-81, \quad \text { hence } \quad v=9+u
$$

Now, considering the equality of the right-hand sides of equations (1) and (2), we can write

$$
\frac{1}{u-15}+\frac{1}{v+6}=\frac{1}{u}+\frac{1}{v}
$$

Bringing to a common denominator:

$$
\frac{u+v-9}{(u-15)(v+6)}=\frac{u+v}{u v}
$$

After rearranging and performing the operations, substituting $v=9+u$, we get $u^{2}+50 u+225=0$. From this, $u_{1}=-5$, $v_{1}=9+u_{1}=4, x_{1}=u_{1}-15=-20, y_{1}=v_{1}+6=10$ and $z_{1}=20 ; u_{2}=-45, v_{2}=-36, x_{2}=-60, y_{2}=-30$, $z_{2}=-20$.

All values can be substituted into the equations, and by substitution, we can verify that the roots indeed satisfy the system of equations.