Let rectangle be $A B C D$.


Here line $A B \| C D$
$\Rightarrow$ slope of line $A B=$ slope of line $C D$
$\Rightarrow \quad-3=\frac{-b}{2} \Rightarrow b=6$
Now line $A B \perp^r B C$
$\Rightarrow($ slope of line $A B) \times($ slope of line $B C)=-1$
$\Rightarrow \quad(-3)\left(\frac{1}{a}\right)=-1 \Rightarrow a=3$
$\therefore$ Eq. of $B C: x-3 y-10=0$
Eq. of $C D: 6 x+2 y+9=0$
$\therefore$ Distance between $A B$ and $C D$
$A B: 3 x+y-4=0 \Rightarrow C D: 3 x+y+\frac{9}{2}=0$
$\therefore$ i.e., $\quad B C=\left|\frac{-4-9 / 2}{\sqrt{9+1}}\right|=\frac{17}{2 \sqrt{10}}$
Now line $A D$ pass through the point $(1,2)$
$\Rightarrow \text { Distance } C D =\text { distance of point }(1,2) \text { from line } B C$
$=\left|\frac{1-6-10}{\sqrt{1+9}}\right|=\frac{15}{\sqrt{10}}$
Now, area of rectangle $A B C D=B C \times C D$
$=\frac{17}{2 \sqrt{10}} \times \frac{15}{\sqrt{0}}=\frac{17 \times 15}{2 \times 10}=\frac{51}{4}$