$$-11 \equiv -11+16 \equiv 5 \equiv 1 \pmod 4$$
So,the congruence becomes:
$$x^2+y^2 \equiv 3 \pmod 4$$
$$x \equiv 0 \pmod 4 \Rightarrow x^2 \equiv 0 \pmod 4$$
$$x \equiv 1 \pmod 4 \Rightarrow x^2 \equiv 1 \pmod 4$$
$$x \equiv 2 \pmod 4 \Rightarrow x^2 \equiv 0\pmod 4$$
$$x \equiv 3 \pmod 4 \Rightarrow x^2 \equiv 1 \pmod 4$$
$$$$
$$y \equiv 0 \pmod 4 \Rightarrow y^2 \equiv 0 \pmod 4$$
$$y \equiv 1 \pmod 4 \Rightarrow y^2 \equiv 1 \pmod 4$$
$$y \equiv 2 \pmod 4 \Rightarrow y^2 \equiv 0\pmod 4$$
$$y \equiv 3 \pmod 4 \Rightarrow y^2 \equiv 1 \pmod 4$$
We can see that it cannot be $x^2+y^2 \equiv 3 \pmod 4$
EDIT:
$$-11 \equiv 1 \pmod 3$$
So,the congruence becomes:
$$x^2+y^2 \equiv 0 \pmod 3$$
$$x \equiv 0 \pmod 3 \Rightarrow x^2 \equiv 0 \pmod 3$$
$$x \equiv 1 \pmod 3 \Rightarrow x^2 \equiv 1 \pmod 3$$
$$x \equiv 2 \pmod 3 \Rightarrow x^2 \equiv 1 \pmod 3$$
$$$$
$$y \equiv 0 \pmod 3 \Rightarrow y^2 \equiv 0 \pmod 3$$
$$y \equiv 1 \pmod 3 \Rightarrow y^2 \equiv 1 \pmod 3$$
$$y \equiv 2 \pmod 3 \Rightarrow y^2 \equiv 1 \pmod 3$$
We can see that $x^2+y^2 \equiv 0 \pmod 3$,only if $x \equiv 0 \pmod 3 \text{ AND } y \equiv 0 \pmod 3$
Notice that if you multiply by $5$ on both sides, you can get rid of the $200$ very quickly and save yourself the trouble of having to use the extended Euclidean algorithm. Also, your mistake is not having a negative sign on both sides of your congruence.
$200x \equiv 13 \pmod{1001} \Rightarrow 1000x \equiv 65 \pmod{1001} \Rightarrow -x \equiv 65 \pmod{1001}$.
Now, multiply both sides by $-1$:
$x \equiv -65 \pmod{1000} \equiv 936 \pmod{1001}$
Best Answer
You can plug $x=12a+2$ into the second and third equations: $$\begin{cases}12a+2\equiv 5 \pmod{15} \\ 12a+2\equiv 11 \pmod{21}\end{cases} \Rightarrow \begin{cases}a\equiv 4 \pmod{5} \\ a\equiv 6 \pmod{7}\end{cases}$$ Now plug $a=5b+4$ into the last equation: $$5b+4\equiv 6 \pmod{7} \Rightarrow b\equiv 6 \pmod{7} \Rightarrow b=7c+6$$ Now go back: $$a=5b+4=5(7c+6)+4=35c+34;\\ x=12a+2=12(35c+34)+34=420c+410 \Rightarrow \\ x\equiv 410 \pmod{420}.$$ Verifying $x=410$: $$11\cdot 410\equiv 4510 \equiv 375\cdot 12+10\equiv 10 \pmod{12};\\ 14\cdot 410\equiv 5740 \equiv 382\cdot 15+10\equiv 10 \pmod{15};\\ 20\cdot 410\equiv 8200 \equiv 390\cdot 21+10\equiv 10 \pmod{21}.$$ You can check $x=420\cdot 2+410=1250$ youserlf.