any solution?

in the part $a)$ it is easy to see that there are no solutions. It is claear that $[a, b, c]$ is a multiple of $[a, b]$, $[b, c]$ and $[c, a]$. Dividing for $ [a, b, c]$ we have a sum of the form $$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1.$$Clearly, the only triples that satisfy are $(3, 3, 3)$, $(2, 3, 6)$ and $(2, 4, 4)$ (and their permutations). If $(x, y, z)=(3, 3, 3)$, then $$[a, b]=[b, c]=[c, a]=k \hspace{0.5cm}\text{and}\hspace{0.5cm} [a, b, c]=3k.$$Of the firt equality, we have that $k$ is a multiple of $a$, $b$ and $c$. Therefore, $[a, b, c]\mid k$, then $3k\mid k$, which is a contradiction. In the other cases it is similary concluded that there is a positive integer $k$ that is a multiple of $a$, $b$ and $c$ and that $[a, b, c]=6k$ or $4k$, in both cases we get at a absurdity.

part b ?

part b) $\newline$ let $d$ be an integer such that $d|a,d|b,d|c,a_1=\frac{a}{d},b_1=\frac{b}{d},c_1=\frac{c}{d}$,then all elements of the equation are divided by d:$ \newline [a_1,b_1]=\frac{a_1b_1}{(a_1,b_1)}=\frac{\frac{ab}{d^2}}{\frac{(a,b)}{d}}=\frac{1}{d} \cdot \frac{ab}{d}=\frac{[a,b]}{d} \newline [a_1,b_1,c_1]=[[a_1,b_1],c_1]=[\frac{[a,b]}{d},\frac{c}{d}]=\frac{1}{d}\cdot [a,b,c] \rightarrow [a_1,b_1]+[b_1,c_1]+[c_1,a_1]=[a_1,b_1,c_1] +(a_1,b_1,c_1) \newline $ We divide $a,b,c$ to $(a,b,c)$ to get $(a,b,c)=1$ hence $[a,b]+[b,c]+[c,a]=[a,b,c]+1 \newline (a,b)|a,b \rightarrow (a,b)|[a,b],[b,c],[c,a] since a|[a,b],a|[a,c],b|[b,c] \rightarrow (a,b)|[a,b,c]+1 \rightarrow (a,b)=1 \newline $ similarly$ (a,b)=(b,c)=(c,a)=1$,hence$ [a,b]=ab,[b,c]=bc,[c,a]=ca,[a,b,c]=abc, ab+bc+ca=abc+1 \newline$ without loss of generality $a \geq b \geq c,abc+1=ab+bc+ca \leq 3ab \rightarrow c\leq 2 \newline$ case $c=1: \newline$ $1=a+b,a,b \text{ are natural,contadiction}$ case $c=2: \newline$ $ab+1=2\cdot a+2\cdot b \rightarrow (a-2)(b-2)=3 \rightarrow a=5,b=3$(we assumed $a\geq b$)$\newline $ so the answer is the form $(a,b,c)=(5k,3k,2k)$ for all natural $k$.