(a)Let required roots are $3\alpha ,\,\,2\alpha ,\,\,\beta $
( ratio of two roots are $3:2$)
$\therefore \,\,\,\,\,\sum \alpha = 3\alpha + 2\alpha + \beta = \frac{{ – ( – 9)}}{1} = 9$
==> $5\alpha + \beta = 9$ ..…$(i)$
$\sum \alpha \beta = 3\alpha .2\alpha + 2\alpha .\beta + \beta .3\alpha $$ = 14$
==> $5\alpha \beta + 6{\alpha ^2} = 14$ …..$(ii)$
and $\sum \alpha \beta \gamma = 3\alpha .2\alpha .\beta = – 24$
==> $6{\alpha ^2}\beta = – 24$ or ${\alpha ^2}\beta = – 4$ …..$(iii)$
from $(i)$,$\beta = 9 – 5\alpha ,$put the value of $\beta $ in $(ii)$
==> $5\alpha (9 – 5\alpha ) + 6{\alpha ^2} = 14$
==>$19{\alpha ^2} – 45\alpha  + 14 = 0$

==>$(\alpha  – 2)(19\alpha  – 7) = 0$

$\alpha  = 2$ or $\frac{7}{{19}}$

$\therefore $ from $(i)$ ,  if $\alpha  = 2,$ then$\beta  = 9 – 5 \times 2$ $= -1$

$\alpha  = 2,\beta  =  – 1$  satisfy the equation $(iii)$ so required  roots are $6, 4, -1.$