3. Let positive integers $a, b, c$ be such that any two of them are not coprime, and $a+b+c$ is a prime number. Find the minimum value of $a+b+c$.

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Let $a, b, c$ be positive integers satisfying the given conditions. Clearly, $a, b, c$ are not equal to 1. We will prove that $a, b, c$ each contain at least two different prime factors.
In fact, if $a$ contains only the prime factor $p$, then since $b, c$ are not coprime with $a$, it follows that $b, c$ are both multiples of $p$, and thus $a+b+c$ is also a multiple of $p$ and greater than $p$, which contradicts the fact that $a+b+c$ is a prime number. Similarly, $b, c$ also have this property.

Take $a=2 \times 3, b=2 \times 5, c=3 \times 5$, in this case $a, b, c$ are pairwise not coprime, and $a+b+c=31$ is a prime number, satisfying the conditions. We will now prove that 31 is the minimum value.
Assume without loss of generality that $a \leqslant b \leqslant c$, then $a \leqslant \frac{a+b+c}{3} \leqslant \frac{29}{3}<10$, and $a$ must contain at least two prime factors, so it can only be $a=2 \times 3=6$.
Note that one of $b, c$ must be odd (otherwise $a+b+c$ would be even), let's assume it is $c$.
Also, at least one of $b, c$ is not a multiple of 3. If $c$ is a multiple of 3, then $c \geqslant 3 \times 5$, in this case $b \geqslant 2 \times 5 \Rightarrow a+b+c \geqslant 31$, a contradiction;
If $c$ is not a multiple of 3, then $c \geqslant 5 \times 7 \Rightarrow a+b+c \geqslant 31$, a contradiction.
In conclusion, the minimum value of $a+b+c$ is 31.