Let $f'(1)=a$, $f''(2)=b$, $f'''(3)=c$. Then $f(x)=x^3+ax^2+bx+c$.

$f'(x)=3x^2+2ax+b \Rightarrow f'(1)=3+2a+b=a \Rightarrow a+b=-3$ ...(i)

$f''(x)=6x+2a \Rightarrow f''(2)=12+2a=b$ ...(ii)

$f'''(x)=6 \Rightarrow f'''(3)=6=c$ ...(iii)

From (ii): $b=12+2a$. Sub in (i): $a+12+2a=-3\Rightarrow3a=-15\Rightarrow a=-5$, $b=2$, $c=6$.

$f(2)=8+4(-5)+2(2)+6=8-20+4+6=\mathbf{-2}$

Wait: $f(2)=8-20+4+6=-2$. So correct index is 2 (option C: $-2$).

\(\Delta n_g = (1+1)-1 = 1\). \(K_c = \frac{[\mathrm{PCl}_3][\mathrm{Cl}_2]}{[\mathrm{PCl}_5]}\) → units: \((\mathrm{mol}\ \mathrm{dm}^{-3})^2 / (\mathrm{mol}\ \mathrm{dm}^{-3}) = \mathrm{mol}\ \mathrm{dm}^{-3} = \mathrm{dm}^3/\mathrm{mol}\).

• $Na^{+}$ from $NaCl$ and $Na_{2}SO_{4}$ (+1)
• $Ba^{2+}$ from $BaCl_{2}$ (+2)
• $Al^{3+}$ from $AlCl_{3}$ (+3)

Subtract equation 1 from equation 2: $(a-1)y = 0$.

Case $a\neq1$: $y=0$. Then from eq.1 and eq.3: $x+z=1$ and $ax+z=0$, giving $(a-1)x=-1$, so $x=\frac{-1}{a-1}$. This always has a solution for $a\neq1$.

Case $a=1$: System becomes $x+y+z=1$, $x+y+z=1$, $x+by+z=0$. The third equation forces $-1=by-... $ i.e. subtracting eq.1: $(b-1)y=-1$. For no solution we need $b=1$.

So $S=\{1\}$ — a single value. Hence $S$ is a singleton.