Before addressing you concerns regarding normalized representatives of congruence classes, I will show you a much simpler way to solve the system of congruences. Using this method I can solve this system mentally in under a minute. So too can you with a little practice.
First, notice that the constant case optimization of CRT applies to the given system, namely $\rm\ x\ \equiv\ 2\ $ mod ($3$ and $\rm\: 7)\ \iff\ 3,\:7\ |\ x-2\ \iff\ 3\cdot 7\ |\ x-2\ \iff\ x\ \equiv\ 2\pmod{21}$
Now to solve $\rm\ x\equiv 2\pmod{21},\ \ x\equiv a\pmod 5\ $ employ Easy CRT to obtain
$$\rm\ x\ \equiv\ 2\ +\ (a-2)\ 21\ \left({21}^{-1}\ mod\ 5\right)\pmod{105}$$
But $\rm\ mod\ 5\!:\ \ 1/21\ \equiv\ 1/1\ \equiv\ 1\ $ so $\rm\ x\ \equiv\ 21\ a - 40\pmod{105}$
In your case $\rm\ a \equiv 3\pmod 5\ $ hence $\rm\ x\ \equiv\ 23\pmod{105}$
Note that $\rm\: -82\equiv 23\pmod{105}\ $ so your solution is indeed correct. If you desire to work only with normalized representatives of congruence classes, e.g. the least nonnegative rep, or the least in absolute value, then you will either need to perform such normalization on your final result, or else work with arithmetic operations that perform such normalizations. It's analogous to the arithmetic of fractions: it doesn't matter whether or not you reduce intermediate results to lowest terms or if, instead, you reduce only the final result to lowest terms. You'll get the same answer either way. But it may be more efficient to reduce intermediate computations to avoid expressions from growing too large. Similarly, here, one can replace any intermediate arithmetic term by any congruent value and it will yield a congruent result. This is the characteristic property of congruences (beyond being equivalence relations).
Hint $\ $ It's a simple constant case of CRT: $\:\!$ if $\rm\:gcd(m,n)=1\:$ then $\rm\:lcm(m,n) = mn,\:$ hence
$\rm\qquad\ \ \ x \equiv a\pmod{m,n}\iff m,n\ |\ x-a\iff mn\ |\ x-a\iff x\equiv a\pmod{mn}$
Thus $\rm\:\!\ \ X+2Y \equiv 3\ \:(mod\ 5,7)\iff X+2Y \equiv 3\pmod{35}\qquad\quad\ \ \ [1]$
and $\rm\ \ \ \: 2X+Y \equiv 4\ \:(mod\ 5,7)\iff 2X+Y \equiv 4\pmod{35}\qquad\quad\ \ \ [2]$
Now $\rm\ 2\cdot [1] - [2]\ \Rightarrow\: 3Y\equiv 2 \equiv -33\ \Rightarrow\ Y \equiv -11\ \Rightarrow\ X \equiv 3-2Y \equiv 25$
Remark $\ $ This simple constant-case optimization of CRT arises quite frequently in practice, esp. for small moduli (by the law of small numbers), so it is well worth checking for. For further examples, see here where it simplified a few page calculation to a few lines, and here and here.
Best Answer
Hint $\ $ Note that by the constant-case optimization of CRT we have
$$\rm\begin{eqnarray} x\equiv 2\pmod 4\\\rm x\equiv 2\pmod 5\end{eqnarray}\ \iff\ x\equiv 2\pmod{4\cdot 5}$$
Now applying Easy CRT to the RHS congruence and the remaining congruence we obtain
$$\rm\begin{eqnarray} x\equiv \color{#C00}2\pmod{\color{blue}{20}}\\\rm x\equiv \color{#0A0}{12}\pmod{27}\end{eqnarray}\ \iff\ x\equiv \color{#0A0}{12}+27\,\left[\dfrac{\color{#C00}2\!-\!\color{#0A0}{12} }{27}\ mod\ \color{blue}{20}\right]\equiv 282\pmod{\color{blue}{20}\cdot 27}$$
where the above fraction is computed as $\rm\, mod\ 20\!:\ \dfrac{10}{27}\equiv \dfrac{10}{7}\equiv\dfrac{30}{21}\equiv \dfrac{10}1,\ $ and $\rm\:-10\equiv 10$