I approached this problem by treating $a$ as a constant and dividing both sides by $a$ to get $bc = 16+\frac{16}{a}b+\frac{16}{a}c$. From this, we get $(b-\frac{16}{a})(c-\frac{16}{a}) = 16+(\frac{16}{a})^2$. To prevent overcounting, we simply set $a$ as the least of the three numbers. Any $a$ greater than $6$ is impossible, as then the left-hand side increases much faster than the right-hand side in the original equation.
Plugging in divisors of 16 for $a$ is easy, so we easily get solutions for $b$ and $c$. However, for non-divisors, it's very annoying to do so (although I already solved the problem). Is there an easier way to find just the number of solutions instead?
This may not be very efficient or fast, but I cannot find anything faster than this method so I will post it and if anyone can make it faster kindly do comment : (inspired by the comment of Jyrki Lahtonen)
Assume $a \leq b \leq c \implies abc = 16(a + b + c) \leq 16(3c) = 48c \implies ab \leq 48$
$\because a \leq b \implies a^2 \leq ab \leq 48 \implies a \leq 6$
Now the only method left is substituting values of a
$a = 1 \implies c = \frac{16 \cdot (b+1)}{b-16} = 16 + \frac{16 \cdot 17}{b - 16}$ Notice that as b increases, c decreases. As $b \leq c$, when we reach a value of b such that at that b, b = c or at the next b, b > c, we stop. This is for all values of a.
Since c is an integer, b - 16 must be a factor of 16*17. Notice 17 is a prime.
Going from smallest values of b ($b > 16 (\because c > 0 \implies 16 + \frac{16 \cdot 17}{b - 16} > 0 \implies \frac{17}{b-16} > -1$ and is b < 16, b - 16 is negative so $17 < -(b-16) \implies 17 < 16 - b \implies b < -1$ but $b > 0$) an this can be proved everytime in such a way so that the denominator should be $> 0$):
b - 16 = 1, b = 17, c = 288 works
b - 16 = 2, b = 18, c = 152 works
b - 16 = 3 is not a factor
b - 16 = 4, b = 20, c = 84,
b - 16 = 5,6,7 are not factors
b - 16 = 8, b = 24, c = 50
b - 16 = 9,10,11,12,13,14,15 are not factors
b - 16 = 16, b = 32, c = 33 works
b - 16 = 17,18,19...31,32,33 are not factors and b - 16 = 34, b = 50, c = 24 < b rejected, stop
$a = 2 \implies c = \frac{8 \cdot (b+2)}{b-8} = 8 + \frac{16 \cdot 5}{b - 8} \;\&\; b > 8$
Going from smallest values of b :
b - 8 = 1, b = 9, c = 88 works
b - 8 = 2, b = 10, c = 48 works
b - 8 = 3 is not a factor
b - 8 = 4, b = 12, c = 28 works
b - 8 = 5, b = 13, c = 24 works
b - 8 = 8, b = 16, c = 18 works
b - 8 = 9 is not a factor and b - 8 = 10, b = 18, c = 16 < b rejected, stop
$a = 3 \implies c = \frac{16 \cdot (b+3)}{3b-16} \;\&\; b > 5$ (here we do not split because there would be no integral splits, it would be sometihing like $\frac{1}{3}$ + something which would achieve nothing)
Going from smallest values of b :
b = 6, c = 72 works
b = 7, c = 32 works
b = 8, c = 22 works
b = 9,10,11 are not factors
b = 12, c = 12 = b stop
$a = 4 \implies c = \frac{4 \cdot (b+4)}{b-4} = 4 + \frac{32}{b - 4} \;\&\; b > 4$
Going from smallest values of b :
b - 4 = 1, b = 5, c = 36 works
b - 4 = 2, b = 6, c = 20 works,
b - 4 = 3 is not a factor
b - 4 = 4, b = 8, c = 12 works and b - 4 > 4 are not factors stop
$a = 5 \implies c = \frac{16 \cdot (b+5)}{5b-16} \;\&\; b \geq a = 5$ (here we do not split because there would be no integral splits, it would be sometihing like $\frac{1}{5}$ + something which would achieve nothing)
Going from smallest values of b :
b = 5 is not a factor and b > 6 denominator becomes > numerator so stop
$a = 6 \implies c = \frac{8 \cdot (b+6)}{3b-8} \;\&\; b \geq a = 6$ (here we do not split because there would be no integral splits, it would be sometihing like $\frac{1}{3}$ + something which would achieve nothing)
Going from smallest values of b :
b = 7, c = 8 and b > 7 denominator becomes > numerator so stop
Thus these are all the values where we can see 17 are $a \neq b \neq c$ (so ordered triplets are $17 \cdot 3! = 102$) and 1 is $a \neq b = c (a = 3, b = c = 12)$ (so ordered triplets are 3)
$\therefore$ total numbered of ordered triplets are 105 and unordered triplets are 18.