From the first proportion, we know that
[tex]3b=8a[/tex]
From the other, we know that
[tex]6c=11b[/tex]
Which implies
[tex]b=\dfrac{8a}{3}=\dfrac{6c}{11}[/tex]
Since [tex]a,b,c[/tex] must be positive integers, [tex]8a[/tex] must be divisible by 3. Since 8 and 3 are coprimes, [tex]a[/tex] itself must be divisible by 3.
Similarly, [tex]6c[/tex] must be divisible by 11. Since 6 and 11 are coprimes, [tex]c[/tex] itself must be divisible by 11.
So, we have [tex]a=3k,\quad c=11n[/tex]. Recalling that
[tex]\dfrac{8a}{3}=\dfrac{6c}{11}[/tex]
we have
[tex]\dfrac{8(3k)}{3}=\dfrac{6(11n)}{11}\iff 8k=6n\iff 4k=3n[/tex]
The smallest possible choice is
[tex]n=4,\quad k=3[/tex]
which implies
[tex]a=3k=9,\quad c=11n=44[/tex]
In fact, this leads to
[tex]b=\dfrac{8\cdot 9}{3}=\dfrac{6\cdot 44}{11}=24[/tex]
So, the three values are
[tex]a=9,\quad b=24,\quad c=44[/tex]
