Find the number of triples $$(a, b, c)$$ of positive integers such that (a) $$ab$$ is a prime; (b) $$bc$$ is a product of two primes; (c) $$abc$$ is not divisible by square of any prime and (d) $$abc\le 30$$.

A prime number has exactly two positive divisors, $$1$$ and itself. Hence the condition $$ab$$ is prime forces
either $$a=1,\;b=p$$ or $$b=1,\;a=p$$ where $$p$$ is some prime.

$$a=1,\;b=p$$ with $$p$$ prime.

The expression $$bc=p\,c$$ must be the product of exactly two primes. Because one prime factor is already $$p$$, $$c$$ itself must be a prime, say $$q$$. If $$q=p$$, then $$bc=p^2$$ and the square of $$p$$ divides $$abc$$, violating condition (c). Therefore $$q\neq p$$.

The size condition is $$abc=1\cdot p\cdot q=pq\le 30.$$

Count all ordered pairs of distinct primes $$(p,q)$$ with $$pq\le 30$$:

$$\begin{aligned} p=2:&\; q=3,5,7,11,13 &\Rightarrow& 5\\ p=3:&\; q=2,5,7 &\Rightarrow& 3\\ p=5:&\; q=2,3 &\Rightarrow& 2\\ p=7:&\; q=2,3 &\Rightarrow& 2\\ p=11:&\; q=2 &\Rightarrow& 1\\ p=13:&\; q=2 &\Rightarrow& 1 \end{aligned}$$

Total for Case 1: $$5+3+2+2+1+1=14$$ triples.

$$b=1,\;a=p$$ with $$p$$ prime.

Now $$bc=c$$ must be the product of exactly two primes. Write $$c=q\,r$$ where $$q,r$$ are primes.

Condition (c) forbids any square factor, so $$q\neq r$$. If $$p$$ coincided with $$q$$ or $$r$$, $$p^2$$ would divide $$abc=pqr$$, again breaking (c). Thus $$p,q,r$$ are three distinct primes.

The size limit gives $$abc=pqr\le 30.$$ The only product of three distinct primes not exceeding 30 is $$2\cdot3\cdot5=30.$$

For the set $$\{2,3,5\}$$ we can choose any of the three primes for $$a=p$$ (because $$b=1$$ is fixed); the remaining two form $$c$$. Hence Case 2 contributes $$3$$ triples.

Adding both cases: $$14+3=17.$$

Therefore, the required number of triples is 17.