Little Cryptogram

We can start from simplifying the cryptogram. $\Large \begin{array} {c c c c c c} & \color{#3D99F6}{Z}& \color{#69047E}{Y}& \color{#D61F06}{X} \\ & & \color{#69047E}{Y} & \color{#D61F06}{X} \\ + & & 0 & \color{#D61F06}{X} \\ \hline\\ \hline & \color{#D61F06}{X} & 0 & \color{#69047E}{Y} \\ \end{array}$ We can get ${\color{#69047E}{Y}}=4$ or $5$ .Consider the column ${\color{#D61F06}{X}}+{\color{#D61F06}{X}}+{\color{#D61F06}{X}}={\color{#69047E}{Y}}$

If ${\color{#69047E}{Y}}=4,{\color{#D61F06}{X}}=8$ ,this fits.

If ${\color{#69047E}{Y}}=5,{\color{#D61F06}{X}}=5$ ,but ${\color{#69047E}{Y}},{\color{#D61F06}{X}}$ are distinct,so this doesn't fit.

So, ${\color{#69047E}{Y}}=4,{\color{#D61F06}{X}}=8,{\color{#3D99F6}{Z}}=7,{\color{#D61F06}{X}}+{\color{#69047E}{Y}}+{\color{#3D99F6}{Z}}=19$

Simplying the cryptogram we find $100z + 20y +3x =100x+y\implies z =\dfrac{97x-19y}{100}$ shows that $z$ is an integer if $97x-19y$ is multiple of $100 <1000$ implying $1<x\leq 9$ . If we further simply the last equation we get $z= \frac{100x-3x - (100-81)y}{100} = x-y + \frac{81y-3x}{100}$ Let $81y-3x = 100n \implies 27y -x =\dfrac{100n}{3} \leq 300$ This now clearly reflects that $3\leq n\leq 9$ . Only possible values we have for $n$ are $3,6,9$ . Note that $y$ is an integer so $1<y\leq 9$ . Thus, either $27y-x = 100$ or $27y-x =200$ . If $27y-x =200$ then $x>10$ . $\therefore 27y-x = 100 \Rightarrow x =27y- 100$ which sure us that $4\leq y\leq 9$ . For $y>4, x\geq 10$ . Hence, only the possible value of $y$ is $4$ . Repluging the value above we find ${\color{#D61F06}x=8}$ and ${\color{#3D99F6}z=7}$ and ${\color{#302B94}y =4}$

Therefore, $x+y+z=\boxed{19}$ .