There are 3 numbers $a,b,c$ such that $HCF(a,b)=l,HCF(b,c)=m$ and $HCF(c,a)=n$. $HCF(l,m)=HCF(l,n)=HCF(n,m)=1$. Find LCM of $a, b, c$. My solution approach for this problem was to use the relationship formula between LCM and HCF of three numbers which is 
$$LCM(p,q,r)=\frac{pqr \times HCF(p,q,r)}{HCF(p,q) \times HCF(q,r) \times HCF(r,p)}$$and upon using this formula I get 
$$LCM(a,b,c)=\frac{abc \times HCF(a,b,c)}{lmn}$$
But I am not able to figure out the the value of $HCF(a,b,c)$.
Can someone please help on this?
Thanks in advance !!!

Let
$$\operatorname{HCF}(a,b,c) = d \tag{1}\label{eq1A}$$
Note $d \mid a$, $d \mid b$ and $d \mid c$. Thus, we also get $d \mid \operatorname{HCF}(a,b) = l$, $d \mid \operatorname{HCF}(b,c) = m$ and $d \mid \operatorname{HCF}(c,a) = n$. Therefore, this then means that $d \mid \operatorname{HCF}(l,m) =  1$ (it also means $d \mid \operatorname{HCF}(l,n) = 1$ and $d \mid \operatorname{HCF}(n,m) = 1$, but this not really needed). This shows that $d = 1$ in \eqref{eq1A}.
With your formula, this then gives
$$\operatorname{LCM}(a,b,c)=\frac{abc \times \operatorname{HCF(a,b,c)}}{lmn} = \frac{abc}{lmn} \tag{2}\label{eq2A}$$
