If $$\overrightarrow p $$ and $$\overrightarrow q $$ are non-collinear unit vectors and $$\left| {\overrightarrow p + \overrightarrow q } \right| = \sqrt 3 ,$$    then $$\left( {2\overrightarrow p - 3\overrightarrow q } \right).\left( {3\overrightarrow p + \overrightarrow q } \right)$$      is equal to :

Solution :
$$\left| {\overrightarrow p + \overrightarrow q } \right| = \sqrt 3 \Rightarrow \overrightarrow {{p^2}} + \overrightarrow {{q^2}} + 2\overrightarrow p \overrightarrow q = 3$$
Since $$\overrightarrow p $$ and $$\overrightarrow q $$ are unit vectors
$$\eqalign{ & {\text{So, }}1 + 1 + 2pq = 3 \cr & \Rightarrow 2pq = 1 \cr & \Rightarrow pq = \frac{1}{2} \cr & \left( {2\overrightarrow p - 3\overrightarrow q } \right)\left( {3\overrightarrow p + \overrightarrow q } \right) = 6\overrightarrow {{p^2}} + 2\overrightarrow p \overrightarrow q - 9\overrightarrow q \overrightarrow p - 3\overrightarrow {{q^2}} = \frac{{ - 1}}{2} \cr} $$