Here is a hands on method of getting the primes you need quite quickly.
Note that $a^{25}-a=a(a^{24}-1)=a(a^{12}+1)(a^6+1)(a^3+1)(a^3-1)$ and $a^{12}+1=(a^4+1)(a^8-a^4+1)$ by successive factorisations using the difference of two squares, so you can obtain a full factorisation for small $a$.
For $a=2$ you get $2^{25}-2=2\cdot17\cdot241\cdot65\cdot9\cdot7$ and the factorisation is easy to complete.
Do a similar thing for $a=3$ to get $3\cdot82\cdot6481\cdot330\cdot 28\cdot 26$
You can then reduce the numbers you have to test to common factors these two, which will be factors of $2\cdot 3\cdot 5\cdot 7\cdot 13$ since no other primes are common.
To prove this works for all factors you need to prove that each of the primes $2,3,5,7,13$ will always divide $a^{25}-a$, and what you need is that $p|a^p-a$ for all primes $p$ and all integers $a$ which is a modest extension of the usual statement of Fermat's little theorem. Use that with the five identified primes.
When solving such questions, it is best to think in terms of prime factorizations instead of integers. So we start by prime-factorizing $360$ and $12$:
\begin{align*}
360 &= 2^3 \cdot 3^2 \cdot 5^1 \\
12 &= 2^2 \cdot 3^1
\end{align*}
$a$ and $b$ must divide $360$, since $360$ is the LCM, so they cannot have any prime factors other than $2, 3, 5$. So
\begin{align*}
a &= 2^x 3^y 5^z \\
b &= 2^{x'} 3^{y'} 5^{z'}
\end{align*}
Now from LCM and GCD being $360$ and $12$, we get
\begin{align*}
\text{max}(x,x') = 3, \text{min}(x,x') = 2 \implies \{x,x'\} = \{2,3\} \\
\text{max}(y,y') = 2, \text{min}(x,x') = 1 \implies \{y,y'\} = \{1,2\} \\
\text{max}(z,z') = 1, \text{min}(z,z') = 0 \implies \{z,z'\} = \{0,1\} \\
\end{align*}
So to write out all possibilities, we just have to assign $2$ and $3$ to $x$ and $x'$, $1$ and $2$ to $y$ and $y'$, and $0$ and $1$ to $z$ and $z'$.
Since we only want unordered possibilities, we can start out taking $z = 0$, $z' = 1$. So all unordered possibilities for $(a,b)$ are, as you wrote:
\begin{align*}
(2^2 3^1, 2^3 3^2 5^1) &= (12, 360) \\
(2^3 3^1, 2^2 3^2 5^1) &= (24, 180) \\
(2^2 3^2, 2^3 3^1 5^1) &= (36, 120) \\
(2^3 3^2, 2^2 3^1 5^1) &= (72, 60). \\
\end{align*}
Best Answer
The equation you're trying to solve for positive integers is
$$a + b + c + \gcd(a,b) + \gcd(b,c) + \gcd(c,a) = \gcd(a,b,c) + \operatorname{lcm}(a,b,c) \tag{1}\label{eq1A}$$
First, note the symmetry in \eqref{eq1A} means that, for any solution $(a,b,c)$, all permutation of the values are also solutions. Next, with $d = \gcd(a, b, c)$, so $a = da'$, $b = db'$ and $c = dc'$, then \eqref{eq1A} becomes
$$\begin{equation}\begin{aligned} da' + db' + dc' + d\gcd(a',b') + d\gcd(b',c') + d\gcd(c',a') & = d + d\operatorname{lcm}(a',b',c') \\ a' + b' + c' + \gcd(a',b') + \gcd(b',c') + \gcd(c',a') & = 1 + \operatorname{lcm}(a',b',c') \end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Since it's also possible to go from the second line \eqref{eq2A} to the first one, this shows we can, WLOG, just solve \eqref{eq1A} with
$$\gcd(a, b, c) = 1 \tag{3}\label{eq3A}$$
Then, from each specific base solution (e.g., where $a \le b \le c$), the other solutions are $(da, db, dc)$ for all integers $d \ge 1$, along with any reordering of the values (e.g., $(dc, da, db)$).
Next, WLOG, have $c$ be the largest value among $a$, $b$ and $c$. From the LHS of \eqref{eq1A} being $\gt c + 1$ and $\lt 6c + 1$, we then also have
$$\operatorname{lcm}(a,b,c) = m_{3}c, \;\; 2 \le m_3 \le 5 \tag{4}\label{eq4A}$$
Note the factors of $m_3$ must come from $a$ and/or $b$, with the rest of the factors of $a$ and $b$ then also being factors of $c$. Thus, using \eqref{eq4A}, we get that
$$\gcd(a,c)=a_1, \;\; \gcd(b,c) = b_1, \;\; a = m_1a_1, \;\; b = m_2b_1, \\ \gcd(a_1,b_1) = 1, \;\; m_3 = \operatorname{lcm}(m_1, m_2) \tag{5}\label{eq5A}$$
WLOG, have $m_1 \le m_2$. Note \eqref{eq5A} is consistent with Dietrich Burde's comment's linked question Prove $\text{lcm}(a,b,c) = \frac{a \cdot b \cdot c \cdot \gcd(a,b,c)}{\gcd(a,b)\gcd(b,c)\gcd(a,c)}$ since $\frac{a\cdot b\cdot c\cdot\gcd(a,b,c)}{\gcd(a,b)\gcd(b,c)\gcd(a,c)} = \frac{m_{1}a_{1}\cdot m_{2}b_{1}\cdot c}{\gcd(m_{1}, m_{2})\cdot a_{1}\cdot b_{1}} = \left(\frac{m_{1}m_{2}}{\gcd(m_{1}, m_{2})}\right)c$. With $m_{1}m_{2} = \operatorname{lcm}(m_{1},m_{2})\gcd(m_{1},m_{2}) \;\;\to\;\; \operatorname{lcm}(m_{1},m_{2}) = \frac{m_{1}m_{2}}{\gcd(m_{1}, m_{2})}$, then $m_3 = \operatorname{lcm}(m_1, m_2)$, as stated in \eqref{eq5A}.
Using \eqref{eq3A}, \eqref{eq4A} and \eqref{eq5A} in \eqref{eq1A} gives that
$$\begin{equation}\begin{aligned} m_1a_1 + m_2b_1 + c + \gcd(m_1,m_2) + b_1 + a_1 & = 1 + m_3c \\ (m_1+1)a_1 + (m_2+1)b_1 & = (m_3-1)c - (\gcd(m_1,m_2) - 1) \\ (m_1+1)a_1 + (m_2+1)b_1 & = (m_3-1)c - e \\ \end{aligned}\end{equation}\tag{6}\label{eq6A}$$
where $e = \gcd(m_1,m_2) - 1$ for simpler algebra. Note that, if $e = 0$, then since $a_1 \mid c$, we have $a_1 \mid (m_2+1)b_1$, but then $\gcd(a_1,b_1)=1$ means $a_1 \mid m_2+1$. Similarly, $b_1 \mid m_1+1$. For a specific example, consider where $m_3 = 2$ in \eqref{eq4A}, with $m_1 = 1$ and $m_2 = 2$ in \eqref{eq5A}. Then \eqref{eq6A} becomes
$$2a_1 + 3b_1 = c \tag{7}\label{eq7A}$$
As discussed above, we have $a_1 \mid 3$ and $b_1 \mid 2$. Checking these cases, while also ensuring the other conditions are met, gives
Thus, only cases #$1$ and #$3$ above work.
With $e \gt 0$, then using
$$c = ja_1b_1, \;\; j \ge 1 \tag{8}\label{eq8A}$$
in \eqref{eq6A} gives that
$$\begin{equation}\begin{aligned} & (m_1+1)a_1 + (m_2+1)b_1 = (m_3-1)ja_1b_1 - e \\ & (m_3-1)j(m_1+1)a_1 + (m_3-1)j(m_2+1)b_1 = (m_3-1)^2j^2a_1b_1 - (m_3-1)je \\ & (m_3-1)^2j^2a_1b_1 - (m_3-1)j(m_1+1)a_1 - (m_3-1)j(m_2+1)b_1 = (m_3-1)je \\ & ((m_3-1)ja_1 - (m_2+1))((m_3-1)jb_1 - (m_1+1)) = (m_3-1)je + (m_2+1)(m_1+1) \end{aligned}\end{equation}\tag{9}\label{eq9A}$$
An example of using this is with the remaining $m_3 = 2$ case where $m_1 = m_2 = 2$, so $e = \gcd(m_1,m_2) - 1 = 1$, with \eqref{eq9A} then becoming
$$(ja_1 - 3)(jb_1 - 3) = j + 9 \tag{10}\label{eq10A}$$
Note that $a = 2a_1$ and $b = 2b_1$ are both even, so $c$ must be odd for $\gcd(a, b, c) = 1$. Thus, $a_1$, $b_1$ and $j$ must also all be odd. Then $ja_1 - 3$ and $jb_1 - 3$ are both even, so $j + 9$ must be a multiple of $4$. The first such case occurs for $j = 3$, with \eqref{eq10A} now being
$$(3a_1 - 3)(3b_1 - 3) = 12 \;\;\; \to \;\;\; (a_1 - 1)(b_1 - 1) = 4 \tag{11}\label{eq11A}$$
The only solution here is $a_1 - 1 = b_1 - 1 = 2 \;\; \to \;\; a_1 = b_1 = 3$, but that contradicts the requirement in \eqref{eq5A} that $\gcd(a_1, b_1) = 1$. The next value to check is $j = 7$, with this resulting in
$$(7a_1 - 3)(7b_1 - 3) = 16 \tag{12}\label{eq12A}$$
Since $7a_1 - 3 \ge 7(1) - 3 = 4$, and similarly for $b_1$, this means the only solution is $a_1 = b_1 = 1$, giving the valid base solution of $(a, b, c) = (2, 2, 7)$.
We don't need to try any larger $j$ values as they require $a_1 \lt 1$ and $b_1 \lt 1$. Another way to see this is that, since $ja_1 - 3$ and $jb_1 - 3$ are strictly increasing functions in $a_1$ and $b_1$, respectively, we get from \eqref{eq10A} that
$$\begin{equation}\begin{aligned} (ja_1 - 3)(jb_1 - 3) & \ge (j - 3)(j - 3) \\ j + 9 & \ge j^2 - 6j + 9 \\ 0 & \ge j^2 - 7j \\ 0 & \ge j(j - 7) \end{aligned}\end{equation}\tag{13}\label{eq13A}$$
which shows $j \le 7$.
This completes the handling for $m_3 = 2$. Dealing with the remaining $m_3$ values, i.e., $3 \le m_3 \le 5$, is similar to what I've shown above.