Prove: Exactly a quarter of 3-part partitions of numbers >2 equal to 0, 2, 10 mod 12 will make a triangle. Consider perimeters $>2$ equal to $0$, $2$, or $10 \mod(12)$. The sequence starts $10, 12, 14, 22, 24, 26, 34, 36, 38, 46, 48, ...$ and we can look at the three part partitions that make triangles.  
$\{\{2, 4, 4\}, \{3, 3, 4\}\}$ out of $8$ partitions for $10$.
$\{\{2, 5, 5\}, \{3, 4, 5\}, \{4, 4, 4\}\}$ out of $12$ partitions for $12$.
$\{\{2, 6, 6\}, \{3, 5, 6\}, \{4, 4, 6\}, \{4, 5, 5\}\}$ out of $16$ partitions for $14$.
$\{\{2, 10, 10\}, \{3, 9, 10\}, \{4, 8, 10\}, \{5, 7, 10\}, \{6, 6, 10\}, \{4, 9, 9\}, \{5, 8, 9\}, \{6, 7, 9\}, \{6, 8, 8\}, \{7, 7, 8\}\}$ out of $40$ partitions for $22$.  
It turns out that for any number $>2$ equal to $0$, $2$, or $10 \mod 12$, exactly one quarter of the $3$-part partitions can make a triangle. It's easy to verify by computer that this true for $n<1000$.  Can someone find a proof? 
Alternately, there is some $n$ not equal to $0$, $2$, or $10 \mod 12$ that also has a quarter of it's partitions making a triangle. What's the first exception?

Let $P_\ell(n)$ denote the number of partitions of $n$ with exactly $\ell$ parts.
The case $n=0\mod 12$:
According to wikipedia, the number of partitions with exactly three parts is equal to
$$P_3(n)=\left[\frac{(n+3)^2}{12}\right]-\left\lfloor \frac n2+1\right\rfloor$$
where $[]$ denotes the nearest integer and $\lfloor\rfloor$ is the lower integer part.
Write $n=12 k$.
Then direct computations give $P_3(12 k)=12 k^2$.
Next notice that any $3$-partition that is not a triangle gives rise to a $2$-partition by summing the two smallest entries. Conversely, any $2$-partition $a+b=n$ with $a\leq b$ gives rise to $\lfloor a/2\rfloor$ $3$-partitions that are not triangles by decomposing $a$ in a $2$-partition. Note that there are precisely $6 k$ such $2$-partitions of $12 k$.
Consequently, the number of $3$-partitions that are not triangles is equal to
$\sum_{i=1}^{6 k} P_2(i)=\sum_{i=1}^{6 k} \lfloor i/2\rfloor=0+1+1+2+2+\dots+(3 k-1)+(3k -1) + 3k=9 k^2 = \frac 34 12 k^2.$
The case $n=-2\mod 12$: the proof works similarly:
$$P_3(12 k-2)=12 k^2-4k,$$
and the number of non-triangles is
$$\sum_{i=1}^{6k-1} \lfloor i/2\rfloor=9 k^2-3k.$$
The case $n=2\mod 12$:
$$P_3(12 k+2)=12 k^2+4k,$$
and the number of non-triangles is
$$\sum_{i=1}^{6k+1} \lfloor i/2\rfloor=9 k^2+3k.$$
