I need to solve $12x-30y+24z=18$ equation as a diophantine equation where $x,y,z$ is whole number I know i can divide with $6$ so I get $2x-5y+4z=3$, I replaced $2x-5y$ with $u$ so I got $u+4z=3$. I see that $u=3$ and $z=0$ is a solution but I got stuck. How can I continue? Is there a way to solve this using extended euclidean algorithm?
Thank you for your help.

Experiments in a spreadsheet show that $x$ $y$ $z$ combinations only add up to a given value if we increment $y$ by $4$ and $z$ by $5$ for a given value of $x$. Further experimentation shows offsets of $1$ and $2$, respectively to make the result zero.
There are infinite solutions starting with $x\in\mathbb{Z}$ and
$$y = 4 n + 2 x + 1,\qquad z = 5 n + 2 x + 2\qquad n \in Z$$
Here are samples with $x\in\{-2,-1,0,1,2,3\}$
For $n=-1$
\begin{equation}
\cdots\quad
(-2,-7,-7)\quad
(-1,-5,-5)\quad
(0,-3,-3)\quad
(1,-1,-1)\quad
(2,1,1)\quad
(3,3,3)\quad\cdots
\end{equation}
For $n=0$
\begin{equation}
\cdots\quad
(-2,-3,-2)\quad
(-1,-1,0)\quad
(0,1,2)\quad
(1,3,4)\quad
(2,5,6)\quad
(3,7,8)\quad\cdots
\end{equation}
For $n=1$
\begin{equation}
\cdots\quad
(-2,1,3)\quad
(-1,3,5)\quad
(0,5,7)\quad
(1,7,9)\quad
(2,9,11)\quad
(3,11,13)\quad\cdots
\end{equation}
for $n=2$
\begin{equation}
\cdots\quad
(-2,5,8)\quad
(-1,7,10)\quad
(0,9,12)\quad
(1,11,14)\quad
(2,13,16)\quad
(3,15,18)\quad\cdots
\end{equation}
For $n=3$
\begin{equation}
\cdots\quad
(-2,9,13)\quad
(-1,11,15)\quad
(0,13,17)\quad
(1,15,19)\quad
(2,17,21)\quad
(3,19,23)\quad\cdots
\end{equation}
and so on.
