Concluding three statements regarding $3$ real numbers. 
$\{a,b,c\}\in \mathbb{R},\ a<b<c,\ a+b+c=6 ,\ ab+bc+ac=9$
Conclusion $I.)\ 1<b<3$
Conclusion $II.)\ 2<a<3$
Conclusion $III.)\ 0<c<1$
Options
By the given statements
$\color{green}{a.)\  \text{Only conclusion $I$ can be derived}}$.
$b.)\ $ Only conclusion $II$ can be derived.
$c.)\ $ Only conclusion $III$ can be derived.
$d.)\ $  Conclusions $I,\ II,\ III$ can be derived.
$e.)\ $  None of the three conclusions can be derived.

$\quad\\~\\$
I tried $(a+b+c)^2=36 \implies a^2+b^2+c^2=18$
and found that $(a,b,c)\rightarrow \{(-1,1,4),(-3,0,3)\}$ satisfies the two conditions
$a^2+b^2+c^2,\ a<b<c $  but not this one $a+b+c=6$
I thought a lot but can't find any suitable pairs.
I look for a simple and short way.
I have studied maths upto $12$th grade.

I shall show that only Conclusion I is correct. For a fixed $b$, we have 
$a+c=6-b$ and $ac=9-(a+c)b=9-(6-b)b=(b-3)^2$.  Hence, the quadratic polynomial 
$x^2-(6-b)x+(b-3)^2$ has two distinct real roots $x=a$ and $x=c$.  Therefore, 
the discriminant $(6-b)^2-4(b-3)^2=3b(4-b)$ of this quadratic is strictly 
positive (hence, $0<b<4$).  Furthermore, the roots of the quadratic are 
$a=\dfrac{(6-b)-\sqrt{3b(4-b)}}{2}$ and $c=\dfrac{(6-b)+\sqrt{3b(4-b)}}{2}$.  As 
$a<b<c$, we must have $\dfrac{(6-b)-\sqrt{3b(4-b)}}{2}<b<\dfrac{(6-b)+\sqrt{3b(4-b)}}{2} $, or equivalently, 
$-\sqrt{3b(4-b)}<3b-6<+\sqrt{3b(4-b)}$.  Hence, 
$\sqrt{3}|b-2|<\sqrt{b(4-b)}$, or $3(b-2)^2<b(4-b)$.  Ergo, $4(b-1)(b-3)<0$. 
That means $1<b<3$.
Note that $b=2$ gives a solution $(a,b,c)=(2-\sqrt{3},2,2+\sqrt{3})$.  
Consequently, Conclusions II and III are false.  In fact, we can prove that 
$0<a<1<b<3<c<4$.  It might be a good exercise for you to show that $0<abc<4$.
