$\gcd(198,288,512)=198x+288y+512z$; $x,y,z\in\Bbb Z$; solve for $x,y,z$

Well, obviously

$$\begin{align} 1 &= \frac{198x + 288y + 512z}{\gcd(198, 288, 512)}\\ \gcd(198, 288, 512)&=\gcd(2\cdot99, 2\cdot144, 2^9) = 2\\ 1 &= 99x + 144y + 256z \end{align}$$

$\gcd(99,144)=9$ so there are $a,b$ where $9 = a99 + b144$
$\gcd(144, 256) = 16$ so there are $c,d$ where $16 = c144 + d256$
$\gcd(9,16) = 1$ so there are $e,f$ where $1 = e9 + 16f$

So we want $$\begin{align} 1&=e(a99 + b144) + f(c144 + d256)\\&= 99ae + 144(be+fc)+ 256df\\ &x = ae;\quad y = be+fc;\quad z = df\\ 1 &= 9e + 16f\end{align}$$

Use Euclid's algorithm: $$\begin{align} 16 &= 9 + 7;& 7&=16-9\\ 9&=7+2;& 2&=9-7\\ 7 &= 3(2) + 1;& 1&= 7-3*2\end{align}$$ $$\begin{align}&=(16-9) - 3(9-7) = (16-9)-3(9 - (16-9))\\ &= 4\cdot16 - 7\cdot9\end{align}$$

So $$e = -7;\quad f= 4\\ x = -7a;\quad y = -7b+4c;\quad z = 4d$$

$$\begin{align} 16 &= c144 + d256\\ 1 &= 9c + 16d\\ 16 &= 9 + 7;\quad 7 = 16 - 9\\ 9 &= 7+2;\quad 2 = 9-7\\ 7 &= 3(2) + 1;\quad 1 = 7-3(2)= (16-9)-3(9-7)\\ &=16 - 4\cdot9 + 3(16-9)\\ &=4\cdot16 - 7\cdot9 \end{align}$$

So $$c= -7;\quad d=4\\ x = -7a;\quad y = -7b-28;\quad z = 16$$

And finally $$\begin{align} 9 &= 99a + 144b\\ 1 &= 11a + 16b\\ 16 &= 11 + 5\\ 11 &= 2\cdot5 + 1\\ 1 &= 11- 2(5)=11-2(16-11) = 3\cdot11 - 2\cdot16 \end{align}$$

So $$a=3;\quad b=-2\\ x = -21;\quad y = -14;\quad z = 16$$

And $$2 = 198(-21) + 288(-14) + 16(512)$$

Cancelling $2$ it suffices to solve $\,\gcd(256,144,99) = 256 z + 144 y + 99 x,\,$ which is straightforward using a version of the extended Euclidean algorithm described here, namely

$$\rm\begin{eqnarray}(1)\quad 256\!\ &=&\ \ 1&\cdot& 256\ +\ \ 0 &\cdot& 144\ +\ 0 &\cdot& 99 \\ (2)\quad 144 \ &=&\ \ 0 &\cdot& 256 \ +\ \ 1 &\cdot& 144\ +\ 0&\cdot& 99\\ (3)\ \,\quad 99 \ &=&\ \ 0 &\cdot& 256 \ +\ \ 0&\cdot& 144\ +\ 1&\cdot& 99\\ 2(2)-(1)\,\rightarrow\, (4)\ \ \quad 32 &=& {-}1&\cdot&256\ +\ \ 2&\cdot&144\ +\ 0&\cdot& 99\\ (3)\!\!-3(4)\,\rightarrow\, (5)\ \ \ \ \quad3 &=&\ \ 3&\cdot&256\ \ -\ 6&\cdot&144\ +\ 1&\cdot&99\\ 11(5)-(4)\,\rightarrow\,(6)\ \ \ \ \quad 1 &\,=\,& 34&\cdot&256 \,-68&\cdot&144+11&\cdot&99 \end{eqnarray}\qquad$$

See here for another worked example, and see here for an explicit formula for the general trivariate linear Diophantine equation (in terms of solutions of associated bivariate equations).