From $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{a}$ we have that $x + y + z \neq 0$ and from $x + y + z = a$ we have $\frac{1}{x + y + z} = \frac{1}{a}$.

Now from $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{a}$ and $\frac{1}{x + y + z} = \frac{1}{a}$ we have $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{x + y + z}$.

$$(xy + yz + zx)(x + y + z) = xyz,$$
$$x^2 y + x y^2 + x y z + x y z + y^2 z + y z^2 + z x^2 + x y z + z^2 x = x y z,$$
$$x^2(y + z) + x y(y + z) + x z(y + z) = 0,$$
$$(y + z)(x^2 + x y + y z + z x) = 0,$$
$$(y + z)[x(x + y) + z(x + y)] = 0,$$
$$(x + y)(y + z)(z + x) = 0.$$
Since $x + y + z = a$ we have $x + y = a - z$, $y + z = a - x$, $z + x = a - y$, and $(a - z)(a - x)(a - y) = 0$. From $(x + y)(y + z)(z + x) = 0$ we obtain that at least one of $a - x, a - y, a - z$ is equal to zero, i.e. $x = a$ or $y = a$ or $z = a$.