Number System , Bases. Need help understanding the solution for this question

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.

Answer 1

$$121x+11y+z=567 $$

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 $$x+y+z = 17$$

Answer 2

Indeed $121x+11y+z = 567$ expresses $(xyz)_{11} = (567)_{10}$, under the restriction that $0 \le x,y,z \le 10$ of course.

Well $121 \times 4 =484 < 567$ while $121\times 5= 605 > 567$ so $x=4$

Then $567-484 = 83$ is left for the two final $x,y$ and so $y=7$ and $83-77=6$ so $z=6$ and so we get $(476)_{11}$ and the sum of the base $11$ "digits" is $17$.