At least one real root is demanded

$x^2 + x + 1 = m , \implies (x + \dfrac{1}{2})^2 + \dfrac{3}{4} = m$

$m \geq \dfrac{3}{4}$

Given expression-

$(m + 1)^2 - (a - 3)(m + 1)m + (a - 4)m^2 = 0$

$2m - 1 - m(a - 3) = 0$

$m = \dfrac{1}{a - 5}$

$\implies a>5$

$m \geq \dfrac{3}{4} , \implies \dfrac{1}{a - 5} \geq \dfrac{3}{4}$

$a \leq \dfrac{19}{3}$

$\boxed{a \in \left(5 , \dfrac{19}{3}\right]}$

only integral value of a = 6 $\implies n =1$

$\dfrac{\displaystyle \sum_{i=1}^n (a_i)^2}{\displaystyle \sum_{j=1}^n (a_j)} \times n$

$= \dfrac{6^2}{6} \times 1 = 6$

It is not too hard to multiply out the brackets and simplify the equation to

$(5-a)x^2+(5-a)x+(6-a)=0$

This has real roots only if $a \neq 5$ and

$(5-a)^2\geq 4(5-a)(6-a)$

Take everything to the left hand side and take out the common factor $(5-a)$ to get

$(5-a)(3a-19)\geq 0$

$\implies5\leq a\leq \dfrac{19}{3}$ which, along with $a\neq 5$

shows that the only possible value is $a=6$ and the result follows easily.

$Let~ Y=X^2+X+1 . ~~\therefore ~ the ~ equation~ is\\(Y+1)^2-(a-3)(Y+1)Y+(a-4)Y^2=0\\\implies~\dfrac{Y+1}{Y}=\dfrac{a-3\pm~\sqrt{(a-3)^2-4(a-4)}}{2}..-tive~ sign~ will~eliminate~a.\\ \therefore~1+\dfrac{1}{Y}= a-4...\implies~Y-\dfrac{1}{a-5}=0...(1).\\Solving~(1)~X=\dfrac{-1~\pm~\sqrt{1-4(1-\dfrac{1}{a-5})}}{2}\\ Given ~condition~\implies~1-4(1-\dfrac{1}{a-5})\ge 0.\\\implies~-3+\dfrac{4}{a-5}\ge 0~~~\implies~a>5. \\Also~\dfrac{a-5}{4}\le \dfrac{1}{3}~~..\implies~a \le \dfrac{19}{3}~~\implies~a\le 6\\\therefore~5 \le a \le 6~and ~required ~value=\dfrac{6^2}{6}*1\\=\boxed{\Large 6}$