Can you find it?

$\begin{aligned} \overline{XX} + \overline{YY} + \overline{ZZ} & = \overline{XYZ} \\ 10X + X + 10Y + Y + 10Z + Z & = 100X+10Y +Z \end{aligned}$

Consider the unit digit on both sides, $X + Y + Z = \overline{1Z} = 10 + Z$ , $\implies X+Y = 10$ .

Now,

$\begin{aligned} 10X + X + 10Y + Y + 10Z + Z & = 100X+10Y +Z \\ 10(X+Y) + X+Y +11Z & = 90X + 10(X+Y) + Z \\ 10 + 11Z & = 90X + Z \\ 10 + 10Z & = 90X \\ Z & = 9X - 1 \end{aligned}$

For $Z \le 9$ , $X = 1$ . And since $X+Y=10$ , $\implies Y=9$ ; and:

$\begin{aligned} 11 + 99 + 11Z & = 190 + Z \\ 110 + 11Z & = 190 + Z \\ 10Z & = 80 \\ \implies Z & = 8 \end{aligned}$

$\implies \overline{XYZ} = \boxed{198}$