A probability problem by Aira Thalca

$x + y + 3z = 36\\ \\ x + y = 36 - 3z$

Now z can take inetgral values from 1 to 12 since x,y,z are non-negative.

$\therefore x + y = 36 , 33, 30, \cdots, 0$ .

So, the possibilites are $\dbinom{37}{1} + \dbinom{34}{1} + \dbinom{31}{1} + \cdots + \dbinom{1}{1}$ .
The series forms an AP with first term 37, common difference -3 and last term 1.

$1 = 37 + (n - 1)(-3)\\ \\ n = 13\\ \\ \therefore \text{sum} = \dfrac{13}{2} \left(1 + 37\right) = \color{#3D99F6}{\boxed{247}}$ .