Let $$\overrightarrow a = {a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k$$

and $$\overrightarrow a \,.\,\left( {3\widehat i - {1 \over 2}\widehat j + 2\widehat k} \right) = 0 \Rightarrow 3{a_1} + {{{a_2}} \over 2} + 2{a_3} = 0$$ ..... (i)

and $$\overrightarrow a \times (2\widehat i + \widehat k) = 2\widehat i - 13\widehat j - 4\widehat k$$

$$ \Rightarrow {a_2}\widehat i + (2{a_3} - {a_1})\widehat j - 2{a_2}\widehat k = 2\widehat i - 13\widehat j - 4\widehat k$$

$$\therefore$$ $${a_2} = 2$$ ..... (ii)

and $${a_1} - 2{a_3} = 13$$ ..... (iii)

From eq. (i) and (iii) : $${a_1} = 3$$ and $${a_3} = - 5$$

$$\therefore$$ $$\overrightarrow a = 3\widehat i + 2\widehat j - 5\widehat k$$

$$\therefore$$ projection of $$\overrightarrow a $$ on $$2\widehat i + 2\widehat j + \widehat k = {{6 + 4 - 5} \over 3} = {5 \over 3}$$