$$
\begin{aligned}
&\text { Given, }{ }^n C_{r-1}=36 \\
&{ }^n C_r=84 \\
&{ }^n C_{r+1}=126 \\
&\text { Divide (i) by (ii), we get }
\end{aligned}
$$
$$
\begin{aligned}
&\Rightarrow \frac{{ }^n C_{r-1}}{{ }^n C_r}=\frac{36}{84} \\
&\Rightarrow \frac{n !}{(r-1) !\{n-(r-1)\} !} \cdot \frac{r !(n-r) !}{n !}=\frac{3}{7} \\
&\Rightarrow \frac{1}{(r-1) !(n-r+1) !} \cdot \frac{r(r-1) !(n-r) !}{1}=\frac{3}{7}
\end{aligned}
$$
$$
\begin{aligned}
&\Rightarrow \frac{1 \cdot r}{(n-r+1)(n-r) !} \cdot(n-r) !=\frac{3}{7} \\
&\Rightarrow \frac{r}{n-r+1}=\frac{3}{7} \\
&\Rightarrow 10 r-3 n=3
\end{aligned}
$$
Divide (ii) by (iii), we get ... (iv)


$$
\begin{aligned}
& \frac{{ }^n C_r}{{ }^n C_{r+1}}=\frac{84}{126} \\
\Rightarrow & \frac{n !}{r !(n-r) !} \cdot \frac{(r+1) !(n-r-1) !}{n !}=\frac{14}{21} \\
\Rightarrow & \frac{1}{r !(n-r)(n-r-1) !} \cdot \frac{(r+1) r !(n-r-1) !}{1}=\frac{2}{3} \\
\Rightarrow & \frac{r+1}{n-r}=\frac{2}{3} \\
\Rightarrow & 3 r+3 \quad=2 n-2 r \Rightarrow 2 n-5 r=3 \quad \ldots(\mathrm{v}) \\
\text { Multiply (iv) by } 2 \text { and (v) by } 3 \text { and add them, we get } \\
20 r-6 n+6 n-15 r=6+9 \\
\Rightarrow & 5 r=15 \Rightarrow r=3 \\
\therefore &{ }^r C_2={ }^3 C_2=\frac{3 !}{2 ! 1 !}=\frac{3 \times 2 !}{2 !}=3
\end{aligned}
$$