On finding triplets that satisfy a certain GCD and LCM property. The question is this:

Find all triplets of positive integers $a,b,c$ satisfying $(a,b,c) = 10$ and $\left[a,b,c\right] = 100$ simultaneously.
Here, $(x,y)$ is the greatest common divisor of $x$ and $y$ and $[x,y]$ is the least common multiple of $x$ and $y$.

It would be very easy if the question was in $2$ variables, because there is a relation between product of the numbers, the GCD and the LCM, but I am not aware of any relation between them when $3$ variables are involved.
For attempting purpose, one can set $a = 10p$, $b = 10q$ and $c = 10r$, where $(p,q,r) = 1$.
Then $[p,q,r] = 10$. Now I don't have an idea to proceed from here.
I am pretty much a beginner to elementary number theory, so might be I have missed something obvious.
Thanks for solutions!
 A: $10=2^15^1$ and $100=2^25^2$. Let $a=2^{a_2}5^{a_5}$, and similarly for $b$ and $c$. Because the gcd expresses the minimum exponent for each prime across its arguments, we have
$$\min(a_2,b_2,c_2)=1\qquad \min(a_5,b_5,c_5)=1$$
Similarly, the lcm expresses the maximum prime exponents, and we have
$$\max(a_2,b_2,c_2)=2\qquad \max(a_5,b_5,c_5)=2$$
One of $a_2,b_2,c_2$ has to be $1$ and another $2$. The third may be either $1$ or $2$. This gives six possibilities for the triple $(a_2,b_2,c_2)$ (as can be directly enumerated easily), and by symmetry there are six possibilities for $(a_5,b_5,c_5)$ independent of the other set of variables. Thus there are $6×6=36$ triples $(a,b,c)$ satisfying the conditions.
A: All three are divisible by $10$, so we divide each by $10$ and find the triples with GCD = $1$ and LCM = $10$.
There can only be factors $2$ and $5$, and both must be there. So, the numbers can be $1$, $2$, $5$ or $10$, and either $2$ and $5$ or just $10$ must be present. They cannot all three be divisible by $2$, or all three divisible by $5$. If we take the numbers in sorted order:
$$
(1,1,10);\  (1,2,5);\  (1,2,10);\  (1,5,10);\  (1,10,10);\  (2,2,5);\  (2,5,5);\  (2,5,10) 
$$
That’s $8$ solutions, and you can arrange the numbers in different order. Multiply by $10$ to solve the original problem.
