Number System , Bases. Need help understanding the solution for this question $$121x+11y+z=567$$
What is $x+y+z$, given $x,y,z$ are digits of a $3$-digit number $(xyz$).
I am stuck at this step:
$$
(xyz)_{11}=(567)_{10}
$$
Is the above step correct or is this correct:
$(xyz)_{11}=(567)_{11}$?
Also please explain the choice.

$$121x+11y+z=567 $$
Working in $\pmod {11}$ we get $z=6$
Thus $$121x+11y =561$$
Divide by $11$ to get $$11x+y=51$$
Working in $\pmod {11}$ we get $y=7$
Plug in the original equation to get $x=4$
As a result $$x+y+z = 17$$
