The position vector of a point P is $\vec{r}=x \vec{i}+y \vec{j}+z \vec{k}$, where $x, y, z \in N$ and $\vec{a}=\vec{i}+\vec{j}+\vec{k}$. If $\vec{r} . \vec{a} = 10$, then the number of possible positions of P is

${}^9C_2$

Given $\vec{r} . \vec{a}=10 \Rightarrow x+y+z=10, x, y, z \geq 1$

The number of possible positions of P

= coefficient of  $x^{10}$  in  $\left(x+x^2+x^3+...\right)^3 $

= coefficient of  $x^7$  in  $(1-x)^{-3}$

$={ }^{3+7-1} C_7={ }^9 C_7={ }^9 C_2=36$

Hence (4) is correct answer.