Given, $f(x)=e x^{2}+b x+c, g(x)=p x^{2}+q x+r$
since, $f(1)=g(1)$
$\Rightarrow \quad a+b+c=p+q+r$
$f(2)=8(2)$
$\Rightarrow \quad 4 a+2 b+c=4 p+2 q+r$
Subtracting Eq.
(tii) from Eq. (i), we get $3 a+b=3 p+q$
$f(3)-g(3)=2$
$\Rightarrow(9 a+3 b+c)-(9 p+3 q+r)=2$
$\Rightarrow \quad 3(3 a+b)+c-3(3 p+q)-r=2$
$\Rightarrow$
$c-r=2 \quad \ldots \text { (iv) }$
$(3 a+b=3 p+q)$
From Eq. (i), $(a-p)+(b-q)+(c-r)=0$
$\Rightarrow \quad(a-p)+(b-q)+2=0$
From Eq. (ii), $4(a-p)+2(b-q)+c-r=0$
$\Rightarrow \quad 2(a-p)+(b-q)+1=0$
Subtracting Eq.(v) from Eq. (vi), we get $(a-p)-1=0$
$a-p=1$
$\therefore$ From Eq. $(v), \quad b-q=-3$
Now, $f(4)-g(4)=(16 a+4 b+c)-(16 p+4 q+r)$
$=16(a-p)+4(b-q)+(c-r) \ldots(v$
Substituting the values of $(a-p),(b-q),(c-r)$ from above in Eq. (vii). we get $\begin{aligned} f(4)-g(4) &=16 \times 1+4(-3)+2 \\ &=16-12+2=6 \end{aligned}$