Find the real value of a for which

$3 i^{3}-2 a i^{2}+(1-a) i+5$

$=-3 i+2 a+(1-a) i+5$

$=(2 a+5)+i(1-a-3)$

$=(2 a+5)+i(-2-a)$

Since, $3 i^{3}-2 a i^{2}+(1-a) i+5$ is real.

$\therefore \operatorname{Im}\left[3 i^{3}-2 a i^{2}+(1-a) i+5\right]=0$

$\Rightarrow-2-a=0$

$\Rightarrow a=-2$

Hence, the real value of $a$ for which $3 i^{3}-2 a i^{2}+(1-a) i+5$ is real is $-2$.