If the equation $2x+2y+z=n$ has $28$ solutions, find the possible values of $n$ 
Let $n$ be a positive integer. If the equation $2x+2y+z=n$ has $28$ solutions, find the possible values of $n$.

I tried taking $2x$ as $a$, and then $2y$ as $b$, and then finding the possibilities. However, I am not sure about the approach I've used.
A little hint would be appreciated.
$x$, $ y$ and $z$ belong to whole numbers.

Assume $x,y,z$ are required to be positive integers.

Consider two cases . . .

Case $(1)$:$\;n$ is even.

From the equation $2x+2y+z=n$, it follows that $z$ is also even.

Writing $z=2w$, and $n=2m$, for some positive integers $w,m$, we get
\begin{align*}
&2x+2y+z=n\\[4pt]
\iff\;&2x+2y+2w=2m\\[4pt]
\iff\;&x+y+w=m\\[4pt]
\end{align*}
By the Stars-and-Bars formula, the equation 
$$x+y+w=m$$
has exactly ${\large{\binom{m-1}{2}}}$ positive integer solution triples $(x,y,w)$, for any given positive integer value of $m$, hence
\begin{align*}
&{\small{\binom{m-1}{2}}}=28\\[4pt]
\implies\;\;&{\small{\frac{(m-1)(m-2)}{2}}}=28\\[4pt]
\implies\;\;&m^2-3m+2=56\\[4pt]
\implies\;\;&m^2-3m-54=0\\[4pt]
\implies\;\;&(m-9)(m+6)=0\\[4pt]
\implies\;\;&m=9\\[4pt]
\implies\;\;&n=18\\[4pt]
\end{align*}

Case $(2)$:$\;n$ is odd.

From the equation $2x+2y+z=n$, it follows that $z$ is also odd.

Writing $z=2w-1$, and $n=2m-1$, for some positive integers $w,m$, we get
\begin{align*}
&2x+2y+z=n\\[4pt]
\iff\;&2x+2y+(2w-1)=2m-1\\[4pt]
\iff\;&x+y+w=m\\[4pt]
\end{align*}
As for case $(1)$, since $m$ is a positive integer such that the equation
$$
x+y+w=m
\qquad\qquad
$$
has exactly $28$ positive integer solution triples $(x,y,w)$, it follows that $m=9$, hence $n=2m-1=17$.

Note:$\;$If $x,y,z$ are only required to be nonnegative integers, then using arguments analogous to the ones given above, we get $n=12$ or $n=11$.
