Let given equation of lines are
AB : 3x - y - 3 = 0   ...(i)
BC: 2x + y - 12 = 0    ...(ii)
CA: x - 2y - 1 = 0  ..(iii)
Solving equation (i) and (ii), we get
x = 3, y = 6 ⇒ B(3, 6)
Solving equation (ii) and (iii), we get
x = 5, Y = 2 ⇒ C (5, 2)
Solving equation (i) and (iii), we get
x = 1, y = 0 ⇒ A (1, 0)

Required area of ΔABC = Area of ΔABP + Area of trapezium BCQP- Area of ΔACQ
=   \(\int _{ AB }^{ }{ y\quad dx } +\int _{ BC }^{ }{ y\quad dx } -\int _{ AC }^{ }{ y\quad dx } \)  
=  \(\int _{ 1 }^{ 3 }{ (3x-3) } dx+\int _{ 3 }^{ 5 }{ (12-2x) } dx-\int _{ 1 }^{ 5 }{ \frac { 1 }{ 2 } } (x-1)dx\)
\(=3{ \left[ \frac { { x }^{ 2 } }{ 2 } -x \right] }_{ 1 }^{ 3 }+{ \left[ 12x-{ x }^{ 2 } \right] }_{ 3 }^{ 5 }-\frac { 1 }{ 2 } { \left[ \frac { { x }^{ 2 } }{ 2 } -x \right] }_{ 1 }^{ 5 }\)
= \(3\left[ \left( \frac { 9 }{ 2 } -3 \right) -\left( \frac { 1 }{ 2 } -1 \right) \right] +[(60-25)-(36-9)]-\frac { 1 }{ 2 } \left[ \left( \frac { 25 }{ 2 } -5 \right) -\left( \frac { 1 }{ 2 } -1 \right) \right] \)
= \(3[2]+[8]-\frac { 1 }{ 2 } [8]\)
= 6 + 8 - 4 = 10 sq. units

Given, x² ≤ y..(i)
and y ≤ |x|...(ii)
Clearly, curve (i) is parabola directed upward with vertex at (0, 0) and symmetrical about y-axis.
Also \(y=|x|=\begin{cases} x\quad if\quad x\ge 0 \\ -x\quad ,if\quad x<\quad 0 \end{cases}\)
The lines y = x and y = - x both passes through origin and have slope of + 1& - 1 respectively.

⇒ Required Area = 2x Standard Area on a side
= \(-\left[ \left( \frac { 1 }{ 2 } -1 \right) -\left( \frac { 16 }{ 2 } -4 \right) \right] \)
= \(2{ \left[ \frac { { x }^{ 2 } }{ 2 } -\frac { { x }^{ 3 } }{ 2 } \right] }_{ 0 }^{ 1 }\)
= \(2\left[ \frac { 1 }{ 2 } -\frac { 1 }{ 3 } \right] =2\times \frac { 1 }{ 6 } \)
= \(\frac { 1 }{ 3 } \) sq.units

Given region {(x,y) : (x² + y² \(\le \) 1 \(\le \) (x + y)}
\(\Rightarrow\) The given region is bounded inside the circle x² + y² = 1 and above the line x + y = 1 as shown in the figure.

\(\therefore\) Required area of the shaded portion
\(=\left| \int _{ circle }^{ }{ ydx } \right| -​​​​\left| \int _{ line }^{ }{ ydx } \right| \)
\(\left| \int _{ 0 }^{ 1 }{ \sqrt { 1-{ x }^{ 2 } } dx } \right| -\left| \int _{ 0 }^{ 1 }{ (1-x)dx } \right| \)
\(=\left| { \left[ \frac { x }{ 2 } \sqrt { 1-{ x }^{ 2 } } +\frac { 1 }{ 2 } { sin }^{ -1 }x \right] }_{ 0 }^{ 1 } \right| -{ \left| \left[ { x-\frac { { x }^{ 2 } }{ 2 } } \right] _{ 0 }^{ 1 } \right| }\)
\(=\left| 0+\frac { 1 }{ 2 } { sin }^{ -1 }(1)-0-0 \right| -\left| 1-\frac { 1 }{ 2 } -0+0 \right| \)
\(=\frac { 1 }{ 2 } \left( \frac { \pi }{ 2 } \right) -\frac { 1 }{ 2 } =\frac { \pi }{ 4 } -\frac { 1 }{ 2 } Sq.units\)

Let R = {(x, y) : y² \(\le \) 4x, 4x² + 4y² \(\le \) 9}
= {(x, y): y² \(\le \) 4x} \(\cap \) {(x, y) : 4x² + 4y² \(\le \) 9]
\(\Rightarrow\) R = R₁ \(\cap \) R₂
where R₁ = {(x, y) : y² \(\le \) 4x}
and R₂ = {(x,y) : 4x² + 4y² \(\le \) 9}
Region R₁: Clearly y² = 4x is the equation of the parabola with vertex at the origin and axis along x-axis. Since we are given that y² \(\le \) 4x, so R₁ is the region lying inside the parabola y² = 4x.
Region R₂ : We have 4x² + 4y² = 9 \(\Rightarrow\) x² + y² = \({ \left( \frac { 3 }{ 2 } \right) }^{ 2 }\)
Clearly, it represents a circle with centre at the origin and radius \(\frac { 3 }{ 2 } \).  It is given that x² + y² \(\le \) \(\frac { 9 }{ 4 } \)
So R₂ is the region lying inside the circle. x² + y² = \({ \left( \frac { 3 }{ 2 } \right) }^{ 2 }\)
Thus the region R is the region bounded by the parabola y² = 4x and the circle x² + y² = (3/2)² as shown by the shaded portion.

and 4x² + 4y² = 9
Put y² = 4x from (i) into (ii), we get
4x² + 16x = 9
\(\Rightarrow\) 4x² + 16x - 9 = 0
\(\Rightarrow\) (2x + 9)(2x - 1) =0
\(\Rightarrow x=\frac { 1 }{ 2 } orx=-\frac { 9 }{ 2 } \)
From (i), x = \(\frac { 1 }{ 2 } \Rightarrow y=\pm \sqrt { 2 } and\quad x=-\frac { 9 }{ 2 } \Rightarrow y\) is imaginary, so the curves intersect at (1/2, \(\sqrt { 2 } \)) and \(\left( \frac { 1 }{ 2 } ,-\sqrt { 2 } \right) \). So both the curves are symmetrical about x-axis.
So required area = 2 (Area of the shaded region lying about x-axis)
Now, Area (OADO)= \(\int _{ 0 }^{ 1/2 }{ { y }_{ 1 } } dx\)
and Area (ADCA) = \(\int _{ 1/2 }^{ 3/2 }{ { y }_{ 2 } } dx\)
\(=\int _{ 1/2 }^{ 3/2 }{ \sqrt { \frac { 9 }{ 4 } -{ x }^{ 2 } } } dx\)
Hence, required area :
A = 2[Area OADO + Area ADCA]
\(=2\int _{ 0 }^{ 1/2 }{ 2\sqrt { x } dx } +2\int _{ 1/2 }^{ 3/2 }{ \sqrt { \frac { 9 }{ 4 } -{ x }^{ 2 }dx } } \)
\(=4\times \frac { 2 }{ 3 } { \left[ { x }^{ 3/2 } \right] }_{ 0 }^{ 1/2 }\)+\(\left[ \left\{ \frac { 9 }{ 4 } { sin }^{ -1 }(1) \right\} -\left\{ \frac { 1 }{ 2 } \sqrt { 2 } +\frac { 1 }{ 2 } .\frac { 9 }{ 4 } { sin }^{ -1 }\left( \frac { 1 }{ 3 } \right) \right\} \right] \)
\(=\frac { 2\sqrt { 2 } }{ 3 } +\left[ \frac { 9 }{ 8 } \pi -\frac { 1 }{ \sqrt { 2 } } -\frac { 9 }{ 8 } { sin }^{ -1 }\left( \frac { 1 }{ 3 } \right) \right] \)
\(=\left[ \frac { \sqrt { 2 } }{ 6 } +\frac { 9\pi }{ 8 } -\frac { 9 }{ 8 } { sin }^{ -1 }\left( \frac { 1 }{ 3 } \right) \right] sq. units\)

Getting the points of intersection as (4, 0), (0, 3).
\(\therefore\) Required area 
\(=\int _{ 0 }^{ 4 }{ \frac { 3 }{ 4 } } \sqrt { 16-{ x }^{ 2 } } dx-\frac { 1 }{ 4 } \int _{ 0 }^{ 4 }{ \left( 12-3x \right) dx } \)
\(=\left[ \frac { 3 }{ 4 } \left[ \frac { x }{ 2 } \sqrt { 16-{ x }^{ 2 } } +8{ sin }^{ -1 }\frac { x }{ 4 } \right] -\frac { 1 }{ 4 } \left( 12x-\frac { { 3x }^{ 2 } }{ 2 } \right) \right] \)
\(=\left( \frac { 3 }{ 4 } .8.\frac { \pi }{ 2 } -6 \right) =\left( 3\pi -6 \right) \ sq.units\)

Given, x² \(\le \) y....(i)
and y \(\le \) |x| ...(ii)
Clearly curve (i) is parabola directed upward with vertex at (0, 0) and symmetrical about y-axis.
Also,\(y=\left| x \right| =\begin{cases} x\quad ,\quad if\quad x\ge 0 \\ -x,\quad if\quad x<0 \end{cases}\)
The lines y = x and y = -x both passes through origin and have slope of +1 & -1 respectively.
\(\Rightarrow\)Required Area = 2 Standard Area on a side 
\(=-\left[ \left( \frac { 1 }{ 2 } -1 \right) -\left( \frac { 16 }{ 2 } -4 \right) \right] \)
\(=2\left[ \frac { { x }^{ 2 } }{ 2 } -\frac { { x }^{ 3 } }{ 3 } \right] _{ 0 }^{ 1 }\)
\(=2\left[ \frac { 1 }{ 2 } -\frac { 1 }{ 3 } \right] =2\times \frac { 1 }{ 6 } \)
\(=\frac { 1 }{ 3 } sq.units\)

The curves y² = 4ax and x² = 4ay intersect at points, where \(\left( \frac { { x }^{ 2 } }{ 4a } \right) ^{ 2 }\)= 4ax

\(\Rightarrow \frac { { x }^{ 4 } }{ 16{ a }^{ 2 } } =4ax\)
\(\Rightarrow { x }^{ 4 }=64{ a }^{ 3 }x\)
\(\Rightarrow x\left( { x }^{ 3 }-64{ a }^{ 3 } \right) =0\)
\(\Rightarrow x=0\ or\ x=4a\)
We plot the curves on same system of axes to get the required region.
\(\therefore\) The enclosed area \(=\int _{ 0 }^{ 4a }{ \left( \sqrt { 4ax } -\frac { { x }^{ 2 } }{ 4a } \right) } \)
\(=\left[ 2\sqrt { a } \frac { 2 }{ 3 } { x }^{ \frac { 3 }{ 2 } }-\frac { { x }^{ 3 } }{ 12\quad a } \right] _{ 0 }^{ 4a }\)
\(=\frac { 4 }{ 3 } \sqrt { a } { \left( 4a \right) }^{ \frac { 3 }{ 2 } }-\frac { { \left( 4a \right) }^{ 3 } }{ 12a } -0\)
\(=\frac { 32{ a }^{ 2 } }{ 3 } -\frac { 16{ a }^{ 2 } }{ 3 } =\frac { { 16a }^{ 2 } }{ 3 } sq.units.\)

The given circles are x² + y² = 9....(i)
and (x-3)² + y² = 9 ...(ii)
These circles intersect at  \(A\left( \frac { 3 }{ 2 } ,\frac { 3\sqrt { 3 } }{ 2 } \right) \ and\ B\left( \frac { 3 }{ 2 } ,-\frac { 3\sqrt { 3 } }{ 2 } \right) \)
The area shaded in the figure

\(\therefore\) The required area ​​​\(=2\int _{ 0 }^{ 3/2 }{ { y }_{ 2 }dx+2\int _{ 3/2 }^{ 3 }{ { y }_{ 1 }dx } } \)   (In view of symmetry)
\(=2\int _{ 0 }^{ 3/2 }{ \sqrt { { 3 }^{ 2 }-\left( x-3 \right) ^{ 2 } } } dx+2\int _{ 3/2 }^{ 3 }{ \sqrt { { 3 }^{ 2 }-{ x }^{ 2 } } } dx\)
\(=2\left[ \frac { \left( x-3 \right) \sqrt { 9-{ \left( x-3 \right) }^{ 2 } } }{ 2 } +\frac { 9 }{ 2 } { sin }^{ -1 }\left( \frac { x-3 }{ 3 } \right) \right] _{ 0 }^{ 3/2 }+2\left[ \frac { x\sqrt { 9-{ x }^{ 2 } } }{ 2 } +\frac { 9 }{ 2 } { sin }^{ -1 }\frac { x }{ 3 } \right] _{ 3/2 }\)
\(=\left[ -\frac { 3 }{ 2 } .\frac { 3\sqrt { 3 } }{ 2 } +9{ sin }^{ -1 }\left( -\frac { 1 }{ 2 } \right) -0-9{ sin }^{ -1 }\left( -1 \right) \right] +\left[ 0+9{ sin }^{ -1 }\left( 1 \right) -\frac { 3 }{ 2 } .\frac { 3\sqrt { 3 } }{ 2 } -9{ sin }^{ -1 }\left( \frac { 1 }{ 2 } \right) \right] \)
\(=\frac { -9 }{ 4 } \sqrt { 3 } -9{ sin }^{ -1 }\left( \frac { 1 }{ 2 } \right) +9{ sin }^{ -1 }\left( 1 \right) +9{ sin }^{ -1 }\left( 1 \right) -\frac { 9 }{ 4 } \sqrt { 3 } -9{ sin }^{ -1 }\left( \frac { 1 }{ 2 } \right) \)
\(=-\frac { -9 }{ 2 } \sqrt { 3 } +18{ sin }^{ -1 }\left( 1 \right) -18{ sin }^{ -1 }\left( \frac { 1 }{ 2 } \right) \)
\(=-\frac { 9 }{ 2 } \sqrt { 3 } +18\left( \frac { \pi }{ 2 } -\frac { \pi }{ 6 } \right) \)
\(=6\pi -\frac { 9 }{ 2 } \sqrt { 3 } \ sq.units\)

Eqns. of PQ, QR and PR are:
PQ : y = \(\frac{5}{2}\)(x-2)
QR : y = 9-x
PR : y = \(\frac{3}{4}\) (x-2)
Req. Area = \(=\int _{ 2 }^{ 4 }{ \frac { 5 }{ 2 } \left( x-2 \right) dx+\int _{ 4 }^{ 6 }{ \left( 9-x \right) dx-\int _{ 2 }^{ 6 }{ \frac { 3 }{ 4 } \left( x-2 \right) dx } } } \)
​​​\(=\left[ \frac { 5 }{ 4 } { \left( x-2 \right) }^{ 2 } \right] _{ 2 }^{ 4 }-\frac { 1 }{ 2 } \left[ \left( 9-x \right) ^{ 2 } \right] _{ 4 }^{ 6 }-\frac { 3 }{ 8 } \left[ \left( x-2 \right) ^{ 2 } \right] _{ 2 }\)
= 5 + 8 - 6 = 7 sq.units.

Equation of normal (OP) \(\Rightarrow\) y = \(\sqrt3\) x
Equation of tangent (PQ) is
y- \(\sqrt3\) =\(\frac { 1 }{ \sqrt { 3 } } \) (x - 1)  
\(\Rightarrow\) y = \(\frac { 1 }{ \sqrt { 3 } } \)(4 - x)
Co-ordinates of point Q is (4, 0).
\(\therefore\) Required Area = \(\int _{ 0 }^{ 1 }{ \sqrt { 3 } x\ dx+\int _{ 1 }^{ 4 }{ 4\ \frac { 1 }{ \sqrt { 3 } } \left( 4-x \right) dx } } \)
\(=\sqrt { 3 } \left[ \frac { { x }^{ 2 } }{ 2 } \right] _{ 0 }^{ 1 }+\frac { 1 }{ \sqrt { 3 } } \left[ 4x-\frac { { x }^{ 2 } }{ 2 } \right] _{ 1 }^{ 4 }\)
\(=\frac { \sqrt { 3 } }{ 2 } +\frac { 1 }{ \sqrt { 3 } } \left[ 16-8-4+\frac { 1 }{ 2 } \right] \)
\(=2\sqrt { 3 } \ sq.units\)