$\begin {array} {1 1} (A)\;27 & \quad (B)\;12 \\ (C)\;6 & \quad (D)\;2 \end {array}$

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Since $\overrightarrow r.\overrightarrow a=5$, $x+y+z=5$, where $x,y,z\in N$

$\therefore$ The no. of possible points p is taken from no. of combinations of 3 natural numbers whose sum is 5 along with their arrangements.

$(1,1,3)..........3\: arrangements.$

$(2,2,1)..........3\: arrangements.$

$\therefore$ The number of possible points = 6

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