The position vector of a point $P$ is given by $\overrightarrow r=x\hat i+y\hat j+z\hat k$, where $x,y,z\in N$. Position vector of point A is given by $\overrightarrow a=\hat i+\hat j+\hat k$. How many points P are possible so that $\overrightarrow r.\overrightarrow a=5$ ?
$\begin {array} {1 1} (A)\;27 & \quad (B)\;12 \\ (C)\;6 & \quad (D)\;2 \end {array}$