Solution: Let $2x + 3y = u.$ Then we must solve $\begin{align} u + 5z = 900 \tag 1 \\ 2x + 3y = u \tag 2 \end{align}$
For $(1),$ a particular solution is $(u_0, z_0) = (0, 180).$ Hence, all the integral solutions of $(1)$ are $\begin{cases} u = 5t \\ z = 180 - t \end{cases} (t \in \mathbb Z)$
Substituting $u = 5t$ into $(2)$ gives $2x + 3y = 5t$ whose particular solution is $(x_0, y_0) = (t, t).$ Hence all the integral solutions of $(2)$ are $\begin{cases} x = t - 3s \\ y = t + 2s \end{cases} (t \in \mathbb Z)$
Thus all the integral solutions of $2x + 3y + 5z = 900$ are given by $$\begin{cases} x = t - 3s \\ y = t + 2s \\ z = 180 - t \end{cases} (s,t \in \mathbb Z)$$
Now suppose $x, y, z \ge 0.$
Note, $180 - t \ge 0 \implies t \le 180$ and so $t + 2s \ge 0 \implies s \ge -90$ and $t - 3s \ge 0 \implies s \le 60.$ Thus we have $-90 \le s \le 60.$
Consider $0 \le s \le 60.$ Now $t \le 180, \ t \ge 3s \implies 3s \le t \le 180$. Thus in this range of $s$, there are $180 - 3s + 1 = 181 - 3s$ of $t$'s.
Consider $-90 \le s < 0.$ Now $t \le 180, \ t \ge -2s \implies -2s \le t \le 180$. Thus in this range of $s$, there are $180 + 2s + 1 = 181 + 2s$ of $t$'s.
Range $0 \le s \le 60$ has the following points:
$(0, 181 - 3(0)), \ (1, 181 - 3(1)), (2, 181 - 3(2), \ldots (60, 181 - 3(60))$ of which there are $61.$
The range $-90 \le s < 0$ must have $91$ points. In sum, we have $61 + 91 = 152$ points for $x, y, z \ge 0.$
My question:
According to the book the answer is $\displaystyle{\sum_{s = 0}^{60}(181 - 3s) + \sum_{s = -90}^{-1}(181 + 2s) = 13651.}$ I don't understand why they took the sum of all $t$'s in the range of $s$. That means some of my denotations and labels above must be incorrect. Where's the mistake? Thanks.
edit:
I think I see my mistake. The number $181 - 3s$ is the number of $t$'s, not necessarily the form of $t$. Given that, the number of ordered pairs (in the given range) must be $(181 - 3s)*61$ by the product rule.
Not an answer, but potentially helpful:
The following MiniZinc model confirms, that there are actually $13651$ solutions.