We have, $x+y+z=15$

$$ \begin{aligned} \text { Total number of solution } & ={ }^{15+3-1} C_{3-1} \\\\ & ={ }^{17} C_2=\frac{17 \times 16}{1 \times 2}=136 \end{aligned} $$

Now, we need to exclude the solutions where two of $(x, y, z)$ are the same.

1) For the case $x = y \neq z$ :

$ 2x + z = 15 $

The solutions are :

$ x = 0, z = 15 $

$ x = 1, z = 13 $

$ x = 2, z = 11 $

$ x = 3, z = 9 $

$ x = 4, z = 7 $

$ x = 5, z = 5 $ (Not valid as all are the same)

$ x = 6, z = 3 $

$ x = 7, z = 1 $

Out of these, 7 are valid.

Similarly, for the cases $y = z \neq x$ and $z = x \neq y$, there will be 7 valid solutions for each, so a total of $ 7 \times 3 = 21 $ solutions where two of the variables are equal.

Thus, the number of triplets where all are distinct is :

$ 136 - 21 = 115 $

There is one solution in which $\mathrm{x}=\mathrm{y}=\mathrm{z}$

Required answer $=136-21-1=114$