It's usually easier to solve the congruences successively, i.e. a pair at a time, replacing a pair of congruences by their solution's congruence. So we take the general solution of the first congruence, substitute it into the second, solve that, then substitute that solution into the third, etc, continually replacing the top $\,2\,$ congruences by an equivalent congruence, as summarized below
$\qquad\quad\begin{align} x\equiv 3\!\!\!\pmod{\! 5}\\ x\equiv 1\!\!\!\pmod{\! 4}\\ x\equiv 2\!\!\!\pmod{\! 3}\end{align}\! $
$\iff \begin{align} \\ x\equiv \color{#0a0}{13}\!\!\!\pmod{\!\color{#0a0}{20}}\\ x\equiv \ \ 2 \!\!\!\pmod{\!3} \ \ \end{align}\!\!$
$\begin{align} \\ \\ \iff x\equiv \color{#c00}{53}\!\!\!\pmod{\!60}\end{align}$
Below is the congruence arithmetic justifying the above reductions.
${\rm mod}\ 5\!:\,\ 3\equiv x\iff x = \color{#90f}{3\! +\! 5\:\!j}.\,$ Substituting this value of $\,x\,$ in the 2nd congruence we get
${\rm mod}\ 4\!:\,\ 1\equiv x = \color{#90f}{3\!+\!5\:\!j}\equiv -1\!+\!j\iff j\equiv 2\iff \color{#0a0}{j=2\!+\!4k}$
Thus the first $2$ congruence are equivalent to $\,x = \color{#90f}{3\!+\!5}\:\!\color{#0a0}j \equiv \color{#0a0}{13\!+\!20k}.\,$ Again, substituting this value of $\,x\,$ into the 3rd congruence yields
${\rm mod}\ 3\!:\,\ 2\equiv x = \color{#0a0}{13+20k}\equiv 1\!-\!k\iff k\equiv 2\iff \color{#c00}{k=2\!+\!3n} $
Therefore the $3$ congruences are equivalent to $\,x = \color{#0a0}{13\!+\!20}\:\!\color{#c00}k = \color{#c00}{53+60n}$
Remark $ $ Below is a summary of the reduction of the top two congruences:
$\qquad \begin{align} x\equiv 3\!\!\!\pmod{\! 5}\\ x\equiv 1\!\!\!\pmod {\!4}\end{align}$
$\iff \begin{align} x = \color{#90f}{3\! +\! 5\color{#0a0}j}\\ \color{#0a0}{j=2\!+\!4k}\end{align}$ $\iff x =\color{#90f}{3\!+\!5}(\color{#0a0}{2\!+\!4k})\overbrace{= \color{#0a0}{13+20k}}^{\large\,\ \equiv\ \ \color{#0a0}{13}\!\pmod{\!\color{#0a0}{\!\!20}}\!}$
Note that all reductions are $\iff$ i.e. "if and only if", i.e. equivalences, because we desire to replace two congruences by an equivalent congruence. This means each reduction step must be reversible. In particular, this means that if we encounter a congruence of the form $\,ac\, x\equiv bc\pmod{\!mc}\,$ then any cancelling of $\,c\neq 0\,$ must also be done to the modulus, i.e.
$$ a\color{#c00}c\,x\equiv b\color{#c00}c\!\!\!\pmod{m\color{#c00}c}\iff mc\mid (ax\!-\!b)c \iff m\mid ax\!-\!b\iff ax\equiv b\!\!\!\pmod m$$
i.e. we must cancel $\,\color{#c00}{c\,\ \rm everywhere}\,$ (compare here for a fractional view of this). Such cancellations occur frequently in the general case when the moduli are not pair-coprime.
Notice we ordered the congruences by largest-moduli-first, so that when the numbers are bigger near the end, the moduli are smaller, so modular arithmetic is easier. This optimization pays off much more when larger moduli are present.
See here for a general formula (Easy CRT) for solving a pair of congruences, including the case of noncoprime moduli.
Galois theory of finite fields sheds some light on this question. In fact, it follows that the latter congruence will always be a consequence of the former! Basically because they are each others Galois conjugates!
As you observed $\alpha^{\pm1}=(n\pm\sqrt{n^2-4})/2$ are the solutions of the quadratic equation
$$
m(x)=x^2-nx+1=0.
$$
Everything takes place in the ring $R$ of algebraic integers of
$\Bbb{Q}(\sqrt{n^2-4})$. From basic algebraic number theory we infer that
$R$ has a prime ideal $\mathfrak{p}$ such that $\mathfrak{p}\cap\Bbb{Z}=p\Bbb{Z}$.
The quotient ring $R/\mathfrak{p}$ is then a finite field $K$. If $n^2-4$ is
a quadratic residue modulo $p$ then $|K|=p$. But if $m(x)$ is irreducible modulo $p$
then $|K|=p^2$.
The idea is that for integers a congruence modulo $p$ is equivalent to congruence modulo $\mathfrak{p}$, and the latter can be decided by projecting everything to the quotient field $K$.
If $m(x)$ factors modulo $p$, then the images of $\alpha^{\pm1}$ in $K$ are residue classes of integers modulo $p$. So Little Fermat says that $\alpha^p\equiv\alpha\pmod{\mathfrak{p}}$ and also $\alpha^{-p}\equiv\alpha^{-1}\pmod{\mathfrak{p}}.$
Consequently
$$\alpha^p+\alpha^{-p}\equiv\alpha+\alpha^{-1}\pmod{\mathfrak{p}}$$
and this implies the same congruence of integers modulo $p$.
If $m(x)$ is irreducible modulo $p$ then its zeros in $K$ are Frobenius conjugates of each other. As those zeros are the projections of $\alpha^{\pm1}$, it follows that
$$
\alpha^p\equiv\alpha^{-1}\pmod{\mathfrak{p}}.
$$
Applying the Frobenius again then gives the congruence
$$
\alpha^p+\alpha^{-p}\equiv\alpha^{-1}+\alpha\pmod{\mathfrak{p}}
$$
and we are done by repeating the earlier argument.
The conclusion is that the condition $$n_1+n_2\equiv n_3\pmod p$$ is all you need for both of the congruences to hold.
Best Answer
The $3,2,1$ are from the right hand side of your congruences.
We know that $x=3$ is a solution to the first congruence, but this doesn't work as a solution to the next 2 congruences. So Chinese remaindering tells you to compute $(3\cdot 5)^{-1}=15^{-1}$ mod $7$. You find that this is $1$ (since $15(1)+(-2)7=1$). Thus $x=3\cdot 15\cdot 1$ ($=3 \cdot 15 \cdot 15^{-1} = 3 \cdot 1 =3$ (mod $7$) is still a solution of $x \equiv 3 \;\mathrm{mod}\; 7$ since $15\cdot 1 \equiv 1 \;\mathrm{mod}\;7$. What is special about this solution? Well, $x=3\cdot 15\cdot 1$ is also congruent to 0 modulo $3$ and $5$ (that's why we were messing with $(3\cdot 5)^{-1}=15^{-1}$).
Next, $x=2$ solves $x\equiv 2\;\mathrm{mod}\;5$, but again does not solve the other two congruences. So you compute $(3\cdot 7)^{-1}=21^{-1}$ mod $5$. You find that this is $1$ (since $21(1)+5(-4)=1$). Thus $x=2\cdot 21\cdot 1$ is still a solution of $x \equiv 2\;\mathrm{mod}\; 5$ while it is also congruent to 0 modulo $3$ and $7$. So now we've found a solution to the second congruence which doesn't interfere with the first and last congruences.
Finally, $x=1$ solves the third congruence but not the first two. So you compute $(5 \cdot 7)^{-1}=35^{-1}$ mod $3$. You find that this is $-1$ (since $35(-1)+3(12)=1$). Thus $x=1\cdot 35 \cdot (-1)$ is still a solution of $x\equiv 1\;\mathrm{mod}\;3$ while it is congruent to $0$ modulo $5$ and $7$.
So now each congruence has a solution which doesn't interfere with the other congruences. Thus adding the solutions together will solve all 3 at the same time.
Therefore, $x = 3\cdot 15\cdot 1 + 2\cdot 21\cdot 1 + 1\cdot 35 \cdot (-1) = 45+42-35=52$ is a solution to all 3 congruences. Since $105$ ($=3\cdot 5 \cdot 7$) is $0$ modulo $3$, $5$, and $7$, adding multiples of $105$ will still leave us with a solution to all $3$ congruences.