An elementary counting argument shows that $2$-$(v,3,3)$ exists only if $v$ is odd (or, more precisely, for $\lambda \equiv 3 \pmod{6}$ a $2$-$(v,3,\lambda)$ exists only if $v \equiv 1 \pmod{2}$). This necessary condition is sufficient.

Arguably the simplest direct construction for the case $\lambda = 3$ that covers all odd $v$ is to use commutative quasigroups. Let $Q = (V, \otimes)$ be a commutative idempotent quasigroup of order $v \equiv 1 \pmod{2}$. A *perpendicular array* of *strength* $3$ is a ${{v+1}\choose{2}}\times 3$ array $L$ whose rows are the ordered triples $(x, y, x\otimes y)$ for $x, y \in V$ and $x \leq y$. Because $Q$ is idempotent, $L$ contains $v$ rows of the form $(x,x,x)$. Deleting such rows and ignoring the ordering on the rows, we obtain a set of triples. Every pair $\{a,b\}$ of distinct elements $a, b \in V$ occurs exactly three times in this set of triples, coming from the ordered triples $(a,b,a\otimes b)$, $(a,c,b)$ when $a \otimes c = c \otimes a = b$, and $(d,b,a)$ when $d\otimes b = b \otimes d =a$. Thus, it suffices to show that a commutative idempontet quasigroup exists for all odd $v = 2n+1$, which is true because we can define $\otimes$ as the binary operation $i\otimes j = (n+1)(i+j) \pmod{v}$ with $V = \{0,1,\dots,v-1\}$.

So, by defining $Q = (V, \otimes)$ as above, the construction is to take $(x,y,x\otimes y)$ for $x\leq y$, throw away $(x,x,x)$, and drop the ordering.