Solution:

The answer is (B).

FIRST SOLUTION

Let $s = x + y + z = 7$, $q = xy + yz + zx$ and $p = xyz = 5$. One can easily notice that $s^{2} = x^{2} + y^{2} + z^{2} + 2(xy + yz + zx) = 27 + 2q$, from which $q = 11$. Let us now consider the polynomial $(t - x)(t - y)(t - z) = t^{3} - (x + y + z)t^{2} + (xy + yz + zx)t - xyz = t^{3} - 7t^{2} + 11t - 5$, which by construction has as its only solutions $x, y$ and $z$. One checks that $p(1) = 1 - 7 + 11 - 5 = 0$ and that $p(5) = 0$, so two of $x, y, z$ are $5$ and $1$, and since the product $xyz$ is $5$ the third variable must be equal to $1$. It follows that $x, y, z$ are $1, 1, 5$ in some order, and hence there are exactly three ordered triples of solutions: $(1, 1, 5)$, $(1, 5, 1)$ and $(5, 1, 1)$.

SECOND SOLUTION

As before we find $11 = x(y + z) + yz$; substituting $y + z = 7 - x$ we thus find $11 = x(7 - x) + yz$. Multiplying both sides by $x$ we get $11x = x^{2}(7 - x) + xyz = -x^{3} + 7x^{2} + 5$, so $x$ is a solution of this equation. One checks that the only solutions are $1$ and $5$, from which the values of $y, z$ are immediately obtained: for example, if $x = 5$, then $x + y = 2$, $xy = 1$, from which $x = y = 1$.