$x+y+x <= 12, \ \ \ \ \ \ \ \ \ \ x,y,z >0$
to solve this problem with less steps we will introduce a dummy variable $w$ which will hold the value $12 - (x+y+z)$ so that RHS of equation becomes stable and we can solve this question easily.
so our new equation will be
$x + y + z + w = 12, \ \ \ x,y,z >0$
now the constrain is $x,y,z>0$
so first we'll calculate all possible solutions then subtract the invalid solutions.
All possible solutions without any restriction = $\large \binom{15}{3}$ = $455$
Now calculating invalid solutions,
when $x$ is $0$,
$y+z+w = 12$
$= \large \binom{14}{2}$
when $y$ is $0$,
$= \large \binom{14}{2}$
when $z$ is $0$,
$= \large \binom{14}{2}$
when both x and y are $0$,
z + w = 12
$= \large \binom{13}{1}$
similarly there are 2 more possibilities (x and z are 0 and z and y are 0),
when x,y and z (all 3) are 0,
$w = 12$
$= \large \binom{12}{0}$
So total invalid solutions = $ \large 3 \times \binom{14}{2} - 3 \binom{13}{1} + 1 \binom{12}{0} = 235$
Total solutions = $455 - 235 =220 $