$15x+72y+21z=9 (1)$
$15+72y = gcd(15,72).r1$
(1) can be written as :
15x+72y=gcd(15,72).r1 (2)
gcd(15,72).r1 + 21z = 9 (3)
$15x+72y=gcd(15,72).r1$
Calculating $GCD(15,72)$ gives:
$72 = 4*15 + 12$
$15 = 1*12 + 3$
$12 = 4*3 + 0$
from (2) and (3) $15x+72y=3.r1 (4)$
$3.r1 + 21z = 9$ (5)
Then applying the Extended Euclidean Algorithm:
$3 = (1 * 15) + (-1 * 12) = (-1 * 72) + (5 * 15)$
A particular solution is $x,y = (5,-1)$ multiply by r1 the $(5r1,-r1)$
general solution $(x,y)=(5r1+24m, -r1-5m)$
$3.r1 + 21z = 9 gcd (3,21) = 3 | 9$ so the is solution $(r1,z) = (3+7n, -n)$
$x=5r1+24m = 5(3+7n)+24m=15+35n+24m$
$y=-r1-5m=-3-7n-5m$
$z=-n$
$(x,y,z) = (15+35n+24m, -3-7n-5m, -n)$
I provide this exercise as an example for diophantine equations with 3 variables however please somebody check for mistakes
I'm not sure what you are after with modular arithmetic but the equation has integer solutions that can be found by other means... assuming a solution in integers is what you seek.
Experimenting with values in a spreadsheet, we can let $x$ be any integer and notice that $y,z$ must jump in respective multiples of $7$ and $24$ if the equation is to sum to a consistent value. Further experimentation locates the constants in their separate equations so that the original sums to zero.
$$15x+72y+21z=9\implies 5x+24y+7z-3=0$$
$$\text{Let}\qquad y = 7 n + 3 x + 1\qquad z = -24 n - 11 x - 3\qquad n,x \in Z$$ Here are some example triples of $(x,y,z)$ as $n$ and $x$ vary from $-2$ to $2$
For $n=-2$ $$\cdots\quad (-2,-19,67)\quad (-1,-16,56)\quad (0,-13,45)\quad (1,-10,34)\quad (2,-7,23)\quad\cdots$$
For $n=-1$ $$ \cdots\quad (-2,-12,43)\quad (-1,-9,32)\quad (0,-6,21)\quad (1,-3,10)\quad (2,0,-1)\quad\cdots$$
For $n=0$ $$\cdots\quad (-2,-5,19)\quad (-1,-2,8)\quad (0,1,-3)\quad (1,4,-14)\quad (2,7,-25)\quad\cdots$$
For $n=1$ $$ \cdots\quad (-2,2,-5)\quad (-1,5,-16)\quad (0,8,-27)\quad (1,11,-38)\quad (2,14,-49)\quad\cdots $$
For $n=2$
$$\cdots\quad (-2,9,-29)\quad (-1,12,-40)\quad (0,15,-51)\quad (1,18,-62)\quad (2,21,-73)\quad\cdots$$