Hint $\rm\ \ x,y,z\:|\:a\!+\!D \iff m = lcm(x,y,z)\:|\:a\!+\!D.\:$ Thus $\rm\:a\equiv -D\pmod m,\:$ which has least natural representative $\rm\:a = m\!-\!D\:$ if $\rm\:0\le D < m.$
Thus, in your example you have that $\rm\:a\:$ is congruent to a constant value mod all moduli, i.e. $\rm\:a \equiv -D \equiv -2\:\ mod\ 3,4,5,\:$ i.e. $\rm\:3,4,5\:|\:a+2\:$ $\Rightarrow$ $\rm\:60 = lcm(3,4,5)\:|\:a + 2,\:$ therefore $\rm\: a = -2 + 60\:\!k,\:$ with least natural solution $\rm\:a = 58.$
You can find much further discussion of this constant-case of CRT (Chinese Remainder) in many of my prior posts, e.g. here.
The second example isn't a constant case of CRT, but one can use an easy form of CRT, viz.
Theorem (Easy CRT) $\rm\ \ $ If $\rm\ p,\:q\:$ are coprime integers then $\rm\ p^{-1}\ $ exists $\rm\ (mod\ q)\ \ $ and
$\rm\displaystyle\quad\quad\quad\quad\quad \begin{eqnarray}\rm n&\equiv&\rm\ a\:\ (mod\ p) \\
\rm n&\equiv&\rm\ b\:\ (mod\ q)\end{eqnarray} \ \iff\ \ n\ \equiv\ a + p\ \bigg[\frac{b-a}{p}\ mod\ q\:\bigg]\ \ (mod\ p\:\!q)$
Proof $\rm\ (\Leftarrow)\ \ \ mod\ p\!:\:\ n\equiv a + p\ (\cdots)\equiv a\:,\ $ and $\rm\ mod\ q\!:\:\ n\equiv a + (b-a)\ p/p \equiv b\:.$
$\rm\ (\Rightarrow)\ \ $ The solution is unique $\rm\ (mod\ p\!\:q)\ $ since if $\rm\ x',\:x\ $ are solutions then $\rm\ x'\equiv x\ $ mod $\rm\:p,q\:$ therefore $\rm\ p,\:q\ |\ x'-x\ \Rightarrow\ p\!\:q\ |\ x'-x\ \ $ since $\rm\ \:p,\:q\:$ coprime $\rm\:\Rightarrow\ lcm(p,q) = p\!\:q\:.\quad$ QED
Applying this to your example we find
$\rm\displaystyle\quad\quad\quad\quad\quad \begin{eqnarray}\rm n&\equiv&\rm\ 2\:\ (mod\ 7) \\
\rm n&\equiv&\rm\ 3\:\ (mod\ 5)\end{eqnarray} \ \iff\ \ n\ \equiv\ 2 + 7\ \bigg[\frac{3-2}{7}\ mod\ 5\:\bigg]\ \ (mod\ 7\cdot 5)$
But $\rm\displaystyle\ mod\ 5\!:\ \frac{1}{7} \equiv \frac{6}2\equiv 3,\: $ therefore $\rm\:\ n\:\equiv\: 2 + 7\cdot 3\equiv 23\pmod{7\cdot 5}$
If the three numbers are $a,b,c$, I believe the answer you are looking for would be $\text{gcd}(a-b, b-c)$.
Say numbers are $a_1, a_2, a_3$. Then we have that for some $x$
$$a_i = q_i N + x$$
This implies that $N$ dividies $a_2 - a_1$ and $a_3 - a_2$, and any $k$ which divides both the differences will necessarily leave the same remainder with the three numbers.
Thus you take the greatest common divisor for the two differences.
Best Answer
HINT: $x,y$ have the same reminder when divided by $d$ if and only if $d$ divides $x-y$.
Once you prove this, the claim follows immediately observing that $x,y,z$ have the same reminder when divided by $d$ if and only if all three pair, $x,y$; $x,z$ and $y,z$ have the same remainder when divided by $d$; if and only if $d$ divides $x-y, x-z, y-z$.