If the position vector of a point P is $\vec r=x\hat i+y\hat j+z\hat k$ where $x, y, z ∈N$ and $\vec a$ is a vector given by $\vec a = \hat i +\hat j+\hat k$, then the total number of possible positions of point P for which $\vec r.\vec a = 10$, is

36

$\vec r.\vec a = 10⇒ x + y + z=10$

The total number of possible positions of P is same as the number of solutions of the above equation in the set of natural numbers. The total number of solutions is

${^{10-1}C}_{3-1}={^9C}_2 = 36$.