Solve $|x - 2| + |x - 1| = x - 3$

If $\frac{x^2+ax+3}{x^2+x+1}$ takes all real values for all real values of $x$,then $a$ lies in the interval

For any real value of $x$,if $\frac{11 x^2+12 x+6}{x^2+4 x+2} \notin(a, b]$,then the value of $x$ for which $\frac{11 x^2+12 x+6}{x^2+4 x+2}=b-a+3$ is

For $0 < c < b < a$,let $(a+b-2c)x^2 + (b+c-2a)x + (c+a-2b) = 0$ and $\alpha \neq 1$ be one of its roots. Then,among the two statements:
$(I)$ If $\alpha \in (-1, 0)$,then $b$ cannot be the geometric mean of $a$ and $c$.
$(II)$ If $\alpha \in (0, 1)$,then $b$ may be the geometric mean of $a$ and $c$.

Let $S = \left\{ \frac{a^2+b^2+c^2}{ab+bc+ca} : a, b, c \in \mathbb{R}, ab+bc+ca \neq 0 \right\}$ where $\mathbb{R}$ is the set of real numbers. Then,$S$ equals

If the figure shows the graph of $y = ax^2 + bx + c$,then . . . . . .