Suppose $a,b,c$ are three distinct natural numbers, such that $3ac = 8(a+b)$. Then, the smallest possible value of $3a+2b+c$ is

From $3ac = 8(a+b)$, we can express $b$ in terms of $a$ and $c$:

$3ac = 8a + 8b$

$8b = 3ac - 8a$

$b = \dfrac{a(3c - 8)}{8}$

Since $b$ must be a natural number, $8$ must divide $a(3c - 8)$, and $3c - 8 > 0$ which means $c \geq 3$.

Also, $a$, $b$, $c$ must all be distinct.

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Substituting $b$ into $3a + 2b + c$:

$3a + 2b + c = 3a + \dfrac{2a(3c - 8)}{8} + c = \dfrac{a(4 + 3c)}{4} + c$

To minimise this, we want small values of $a$ and $c$. We try small values of $c$ systematically.

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$c = 3$:

$b = \dfrac{a(1)}{8} = \dfrac{a}{8}$

Smallest valid $a = 8$, giving $b = 1$. All distinct.

$3(8) + 2(1) + 3 = 29$

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$c = 4$:

$b = \dfrac{a(4)}{8} = \dfrac{a}{2}$

Smallest valid $a = 2$, giving $b = 1$. All distinct.

Verification: $3(2)(4) = 24$ and $8(2 + 1) = 24$ ✓

$3(2) + 2(1) + 4 = 12$

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$c = 5$:

$b = \dfrac{7a}{8}$

Smallest valid $a = 8$, giving $b = 7$. All distinct.

$3(8) + 2(7) + 5 = 43$

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$c = 6$:

$b = \dfrac{10a}{8} = \dfrac{5a}{4}$

Smallest valid $a = 4$, giving $b = 5$. All distinct.

$3(4) + 2(5) + 6 = 28$

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As $c$ grows beyond $4$, the values keep increasing. The case $c = 4$, $a = 2$, $b = 1$ gives the smallest result.

The smallest possible value of $3a + 2b + c = \boxed{12}$