CAT 2019 Shift 2 Paper

Sum of roots $=4\mathit{a}+3\mathit{a}=7\mathit{a}=-\mathit{b}$
Or $\mathit{b}=-7\mathit{a}$
Product of roots $=4\mathit{a}\times 3\mathit{a}=\mathit{c}$
Or $\mathit{c}=12{\mathit{a}}^2$
Now, ${\mathit{b}}^2+\mathit{c}={\left(-7\mathit{a}\right)}^2+12{\mathit{a}}^2=61{\mathit{a}}^2$
Comparing the options.
Option 1: $61{\mathit{a}}^2=3721\Rightarrow {\mathit{a}}^2=61,$ clearly a is not an integer.
Option 2: $61{\mathit{a}}^2=549\Rightarrow {\mathit{a}}^2=9,$ we can have $\mathit{a}=-3$ or 3 (an integer)
Option 3: $61{\mathit{a}}^2=427\Rightarrow {\mathit{a}}^2=7,$ clearly a is not an integer.
Option 4: $61{\mathit{a}}^2=361\Rightarrow {\mathit{a}}^2=\frac{361}{61},$ clearly a is not an integer.