In this above question

we apply bodmas rule,

⇒ $3\frac{1}{4}+2+\frac{1}{2}-\left(1+\frac{5}{6}\right)=\frac{(?{)}^2}{10}+1+\frac{5}{12}$

Or, add the integral part and fractional part separately.

$$\begin{gathered} = (3+2-1)+(\frac{1}{4}+\frac{1}{2}-\frac{5}{6}) \\ = \frac{(?{)}^2}{10}+1+\frac{5}{12} \\ = \left(4\right)+\left(\frac{3+6-10}{12}\right)=\frac{(?{)}^2}{10}+1+\frac{5}{12} \\ = (4-1)+\left(-\frac{1}{12}-\frac{5}{12}\right)=\frac{(?{)}^2}{10} \\ =3-\frac{6}{12}=\frac{(?{)}^2}{10} \\ = 3-\frac{1}{2}=\frac{(?{)}^2}{10} \\ = \frac{5}{2}\times 10=(?{)}^2 \\ = 25=(?{)}^2 \\ =\sqrt{25}=? \\ =?=5 \end{gathered}$$

Hence, Option E is correct.