(c) Since $\alpha ,\;\beta ,\;\gamma ,\;\delta $ form an increasing $G.P.$,

so $\alpha \delta = \beta \gamma $where $\alpha < \beta < \gamma < \delta $.

On solving ${x^2} – 3x + a = 0$,

we get $x = \frac{1}{2}(3 \pm \sqrt {9 – 4a} )$.

Also $\alpha < \beta $.

Hence $\alpha = \frac{1}{2}(3 – \sqrt {9 – 4a} ),\;$

$\beta = \frac{1}{2}(3 + \sqrt {9 – 4a} )$

Similarly from ${x^2} – 12x + b = 0$, we get

$\gamma = \frac{1}{2}(12 – \sqrt {144 – 4b} ),\;$

$\delta = \frac{1}{2}(12 + \sqrt {144 – 4b} )$

Substituting these values of $\alpha ,\;\beta ,\;\gamma ,\;\delta $ in $\alpha \delta = \beta \gamma $

and simplifying, we get $(a,\;b) = (2,\;32)$.

Trick : Check the alternates; only $(c)$ satisfies the condition.