Let $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2-4\mathrm{ax}+5{\mathrm{a}}^2-6\mathrm{a}$. Let p be the smallest positive integral value  of 'a'  for  f(x)>0 $\mathrm{\forall }x\in R$ and S be the set of values of ' a' for which range of f(x) is $[-8,\mathrm{\infty })$. Then

$\begin{array}{l}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2-4\mathrm{ax}+5{\mathrm{a}}^2-6\mathrm{a}>0\mathrm{\forall }\mathrm{x}\in \mathrm{R}\\ \Rightarrow \mathrm{D}=16{\mathrm{a}}^2-4\left(5{\mathrm{a}}^2-6\mathrm{a}\right)<0\text{ [becauseCoeff. of }{\mathrm{x}}^2>0\text{ ] }\\ \Rightarrow -4{\mathrm{a}}^2+24\mathrm{a}<0\Rightarrow {\mathrm{a}}^2-6\mathrm{a}>0\\ \Rightarrow \mathrm{a}<0\text{ or }\mathrm{a}>6\end{array}$

Smallest positive value of a = 7
Also $f\left(x\right)=[x-2a{]}^2+5a^2-6a-4a^2$
$\Rightarrow \text{ least }f\left(x\right)=a^2-6a$
Since range $f\left(x\right)\Rightarrow [-8,\mathrm{\infty })$
$\begin{array}{l}\therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{a}}^2-6\mathrm{a}=-8\stackrel{⩾}{\Rightarrow }{\mathrm{a}}^2-6\mathrm{a}+8=0\Rightarrow \mathrm{a}=2,4\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{S}=\{2,4\}\end{array}$