The equation of the plane passing through the intersection of the planes
 \(\vec { r } .\left( 2\hat { i } +6\hat { j } \right) +12=0\) and \(\vec { r } .\left( 3\hat { i } -\hat { j } +4\hat { k } \right) =0\) 
\(\Rightarrow \vec { r } .\left\{ \left( 2+3\lambda \right) \hat { i } +\left( 6-\lambda \right) \hat { j } +4\lambda k\hat { j } \right\} +12=0....(i) \)
The planes are at a unit distance from origin. Therefore, length of the perpendicular from the origin to the plane (i) = 1 unit.
\(\frac { 12 }{ \sqrt { { \left( 2+3\lambda \right) }^{ 2 }+{ \left( 6-\lambda \right) }^{ 2 }+{ 16\lambda }^{ 2 } } } =1\)
\(\Rightarrow 144=\left( 2+3\lambda \right) ^{ 2 }+\left( 6-{ \lambda } \right) ^{ 2 }+16{ \lambda }^{ 2 }\)
\(\Rightarrow 144=40+26{ \lambda }^{ 2 }\)
\(\Rightarrow 26{ \lambda }^{ 2 }=104\)  
\(\Rightarrow { \lambda }^{ 2 }=4\)
\(\Rightarrow \lambda =\pm 2\)
Putting the value of in equation (i), we get \(\vec { r } .\left( 8\hat { i } +4\hat { j } +8\hat { k } \right)+12 =0\)
and \(\vec { r } .\left(-4\hat { i } +8\hat { j } -8\hat { k } \right)+12 =0\)
which are the equations of the required planes.  These equations can also be written as \(\vec { r } .\left( 2\hat { i } +\hat { j } +2\hat { k } \right)+3 =0\) and \(\vec { r } .\left( \hat {- i } +2\hat { j } -2\hat { k } \right)+3=0\)
The above equations can be written in cartesian from as follows: 2x + y + 2z + 3 = 0 and -x + 2y - 2z + 3 = 0

Circles are x² + y² = 1 and (x - 1)² + y² = 1.
centre (0, 0), radius 1 and centre (1, 0), radius 1.
On plotting the two circles we notice, we have to find area of the shaded portion.
Both curves are symmetrical to the x-axis, as both the functions are even with respect to y.
Eliminating y from two equations, we get

\(1-x^{2}=1-(x-1)^{2} \Rightarrow x=\frac{1}{2}\)
\(\therefore \text { Area }=2 \operatorname{ar}(O A B)=2[\operatorname{ar}(O A L)+\operatorname{ar}(L A B)\)
\(\therefore \text { area }=2\left[\int_{0}^{1 / 2} \sqrt{1-(x-1)^{2}} d x+\int_{1 / 2}^{1} \sqrt{1-x^{2}} d x\right] \)
\(=2\left[\left\{\frac{x-1}{2} \sqrt{1-(x-1)^{2}}+\frac{1}{2} \sin ^{-1}\left(\frac{x-1}{1}\right)\right\}_{0}^{1 / 2}\right. \)
\(\left.+\left\{\frac{x}{2} \sqrt{1-x^{2}}+\frac{1}{2} \sin ^{-1} x\right\}_{1 / 2}^{1}\right] \)
\(=2\left[\left\{-\frac{1}{4} \times \frac{\sqrt{3}}{2}+\frac{1}{2} \sin ^{-1}\left(-\frac{1}{2}\right)\right\}-\left\{0+\frac{1}{2} \sin ^{-1}(-1)\right\}\right. \)
\(\left.+\left\{0+\frac{1}{2} \sin ^{-1}(1)\right\}-\left(\frac{1}{4} \times \frac{\sqrt{3}}{2}+\frac{1}{2} \sin ^{-1} \frac{1}{2}\right)\right] \)
\(=2\left[-\frac{\sqrt{3}}{8}-\frac{\pi}{12}+\frac{\pi}{4}+\frac{\pi}{4}-\frac{\sqrt{3}}{8} \div \frac{\pi}{12}\right] \)
\(=2\left[\frac{\pi}{3}-\frac{\sqrt{3}}{4}\right]=\left(\frac{2 \pi}{3}-\frac{\sqrt{3}}{2}\right) \mathrm{sq} \text { units } \)

The given equation of the circle x² + y² = 8x can be expressed as (x – 4)² + y² = 16. Thus, the centre of the circle is (4, 0) and radius is 4. Its intersection with the parabola y² = 4x gives
x² + 4x = 8x
or x² – 4x = 0
or x (x – 4) = 0
or x = 0, x = 4
Thus, the points of intersection of these two curves are O(0, 0) and P(4,4) above the x-axis.
the required area of the region OPQCO included between these two curves above x-axis is
= (area of the region OCPO) + (area of the region PCQP)
\(=\int_{0}^{4} y d x+\int_{4}^{8} y d x \)
\(=2 \int_{0}^{4} \sqrt{x} d x+\int_{4}^{8} \sqrt{4^{2}-(x-4)^{2}} d x \)
\(=2 \times \frac{2}{3}\left[x^{\frac{3}{2}}\right]_{0}^{4}+\int_{0}^{4} \sqrt{4^{2}-t^{2}} d t, \text { where, } x-4=t\)
\(=\frac{32}{3}+\left[\frac{t}{2} \sqrt{4^{2}-t^{2}}+\frac{1}{2} \times 4^{2} \times \sin ^{-1} \frac{t}{4}\right]_{0}^{4}\)
\(=\frac{32}{3}+\left[\frac{4}{2} \times 0+\frac{1}{2} \times 4^{2} \times \sin ^{-1} 1\right]=\frac{32}{3}+\left[0+8 \times \frac{\pi}{2}\right]=\frac{32}{3}+4 \pi=\frac{4}{3}(8+3 \pi)\)

\(\frac{8}{3}sq.units\)

\(\frac { 5 }{ 2 } sq.units\)

\(\pi\)sq.units.

The given curves are:
y² = (x + 1) and y² = - (x - 1)
These two parabolas meet at A (0,1) and B(0,1)

Therefore.area = 2 [ar.(CAO_ + ar. (OAD)]
\(=2\left[ \overset { 0 }{ \underset { -1 }{ \int { } } } \sqrt { x+1 } dx+\overset { 1 }{ \underset { 0 }{ \int { } } } \sqrt { 1-x } dx \right] \)
\(=2\left[ \left[ \frac { { (x+1) }^{ 3/2 } }{ 3/2 } \right] _{ -1 }^{ 0 }+\left[ \frac { { (1-x) }^{ 3/2 } }{ \frac { 3 }{ 2 } (-1) } \right] _{ 0 }^{ 1 } \right] \)
\(=\frac { 4 }{ 3 } \left[ \left[ { (x+1) }^{ 3/2 } \right] _{ -1 }^{ 0 }+\left[ { -(1-x) }^{ 3/2 } \right] _{ 0 }^{ 1 } \right] \)
\(=\frac { 4 }{ 3 } \left[ (1-0)+(-0+1) \right] \)
\(=\frac { 4 }{ 3 } (2)=\frac { 8 }{ 3 } =2\frac { 2 }{ 3 } sq.units\)

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

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 \(=\int _{ 0 }^{ 4a }{ \left( \sqrt { 4ax } -\frac { { x }^{ 2 } }{ 4a } \right) } \)
\(=\left[ 2\sqrt { a } \frac { 2 }{ 3 } { x }^{ \frac { 3 }{ 2 } }-\frac { { x }^{ 3 } }{ 12\quad a } \right] _{ 0 }^{ 4a }\)
\(=\frac { 4 }{ 3 } \sqrt { a } { \left( 4a \right) }^{ \frac { 3 }{ 2 } }-\frac { { \left( 4a \right) }^{ 3 } }{ 12a } -0\)
\(=\frac { 32{ a }^{ 2 } }{ 3 } -\frac { 16{ a }^{ 2 } }{ 3 } =\frac { { 16a }^{ 2 } }{ 3 } sq.units.\)

Given curves are y = 2x....(i)
and x - y = 4...(ii)
Obviously, curve (i) is right handed parabola having vertex at (0, 0) and axis along +ve direction of x-axis while curve (ii) is a straight line.

For intersection point of curve (i) and (ii) (x-4)² = 2x
\(\Rightarrow\) x²-8x + 16 = 2x
\(\Rightarrow\) x²-10x +16 = 0
\(\Rightarrow\) x²-8x-2x+16 = 0
\(\Rightarrow\) x(x-8)-2(x-8) = 0
\(\Rightarrow\) (x-8) (x-2) = 0
\(\Rightarrow\) x = 2, 8
\(\Rightarrow\) y = -2, 4
Intersection points are (2, -2), (8, 4)
Therefore, required area =  Area of shaded region
\(=\int _{ -2 }^{ 4 }{ \left( y+4 \right) dy-\int _{ -2 }^{ 4 }{ \frac { { y }^{ 2 } }{ 2 } } dy } \)
\(=\int _{ -2 }^{ 4 }{ \left( y+4 \right) dy-\int _{ -2 }^{ 4 }{ \frac { { y }^{ 2 } }{ 2 } } dy } \)
\(=\frac { 1 }{ 2 } .\left[ 64-4 \right] -\frac { 1 }{ 6 } \left[ 64+8 \right] \)
\(=30-\frac { 72 }{ 6 } =18\ sq.units\)

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)
\(=2\int _{ 0 }^{ 3/2 }{ \sqrt { { 3 }^{ 2 }-\left( x-3 \right) ^{ 2 } } } dx+2\int _{ 3/2 }^{ 3 }{ \sqrt { { 3 }^{ 2 }-{ x }^{ 2 } } } dx\)
\(=2\left[ \frac { \left( x-3 \right) \sqrt { 9-{ \left( x-3 \right) }^{ 2 } } }{ 2 } +\frac { 9 }{ 2 } { sin }^{ -1 }\left( \frac { x-3 }{ 3 } \right) \right] _{ 0 }^{ 3/2 }+2\left[ \frac { x\sqrt { 9-{ x }^{ 2 } } }{ 2 } +\frac { 9 }{ 2 } { sin }^{ -1 }\frac { x }{ 3 } \right] _{ 3/2 }\)
\(=\left[ -\frac { 3 }{ 2 } .\frac { 3\sqrt { 3 } }{ 2 } +9{ sin }^{ -1 }\left( -\frac { 1 }{ 2 } \right) -0-9{ sin }^{ -1 }\left( -1 \right) \right] +\left[ 0+9{ sin }^{ -1 }\left( 1 \right) -\frac { 3 }{ 2 } .\frac { 3\sqrt { 3 } }{ 2 } -9{ sin }^{ -1 }\left( \frac { 1 }{ 2 } \right) \right] \)
\(=\frac { -9 }{ 4 } \sqrt { 3 } -9{ sin }^{ -1 }\left( \frac { 1 }{ 2 } \right) +9{ sin }^{ -1 }\left( 1 \right) +9{ sin }^{ -1 }\left( 1 \right) -\frac { 9 }{ 4 } \sqrt { 3 } -9{ sin }^{ -1 }\left( \frac { 1 }{ 2 } \right) \)
\(=-\frac { -9 }{ 2 } \sqrt { 3 } +18{ sin }^{ -1 }\left( 1 \right) -18{ sin }^{ -1 }\left( \frac { 1 }{ 2 } \right) \)
\(=-\frac { 9 }{ 2 } \sqrt { 3 } +18\left( \frac { \pi }{ 2 } -\frac { \pi }{ 6 } \right) \)
\(=6\pi -\frac { 9 }{ 2 } \sqrt { 3 } \ sq.units\)

The given curves are
y = x² (parabola)
y = x (line)
These intersect at
O(0, 0) and A(1, 1).
 
The area bounded by the curves = Shaded area
 \(=\int _{ 0 }^{ 1 }{ \left( { y }_{ 2 }-{ y }_{ 1 } \right) } dx\)
\(=\int _{ 0 }^{ 1 }{ \left( x-{ x }^{ 2 } \right) } dx\)
\(=\left[ \frac { { x }^{ 2 } }{ 2 } -\frac { { x }^{ 3 } }{ 3 } \right] _{ 0 }^{ 1 }\)
\(=\frac { 1 }{ 2 } -\frac { 1 }{ 3 } =\frac { 1 }{ 6 } sq.units\)

Equation of : 
AB is : \(y=\frac { 1 }{ 2 } \left( 3x+7 \right) \)
BC is : \(y=\frac { 1 }{ 2 } \left( 11-x \right) \)
AC is : \(y=\frac { 1 }{ 2 } \left( x+5 \right) \)
Required Area
\(=\frac { 1 }{ 2 } \int _{ -1 }^{ 1 }{ \left( 3x+7 \right) } dx+\frac { 1 }{ 2 } \int _{ 1 }^{ 3 }{ \left( 11-x \right) } dx\frac { 1 }{ 2 } \int _{ -1 }^{ 3 }{ \left( x+5 \right) dx } \)
\(=\left[ \frac { 1 }{ 12 } { \left( 3x+7 \right) }^{ 2 } \right] _{ -1 }^{ 1 }-\frac { 1 }{ 4 } \left[ \left( 11-x \right) ^{ 2 } \right] _{ 1 }^{ 3 }-\frac { 1 }{ 4 } \left[ \left( x+5 \right) ^{ 2 } \right] _{ -1 }^{ 3 }\)
\(=7+9-12=4\ sq.units\)

Let the line AB, BC and CA have equations 2x + y = 4, 3x - 2y = 6 and x - 3y + 5 = 0 resp. B(2, 0), C(4, 3) and A(1, 2)

Area \(=\int _{ 1 }^{ 4 }{ \frac { 1 }{ 3 } } \left( x+5 \right) dx-\int _{ 1 }^{ 2 }{ \left( 4-2x \right) } dx-\int _{ 2 }^{ 4 }{ \frac { 1 }{ 2 } } \left( 3x-6 \right) dx\)
\(=\frac { 1 }{ 3 } \left[ \frac { { \left( x+5 \right) }^{ 2 } }{ 2 } \right] _{ 1 }^{ 4 }+2\left[ \frac { { \left( 2-x \right) }^{ 2 } }{ 2 } \right] _{ 1 }^{ 2 }-\frac { 3 }{ 2 } \left[ \frac { { \left( x-2 \right) }^{ 2 } }{ 2 } \right] _{ 2 }^{ 4 }\)
\(=\left( \frac { 81 }{ 6 } -\frac { 36 }{ 6 } \right) +\left( 0-1 \right) -\frac { 3 }{ 4 } .4\)
\(=\frac { 15 }{ 2 } -1-3=\frac { 7 }{ 2 } sq.units.\)

Given 4y = x²
\(\Rightarrow 4\frac { dy }{ dx } =2x\)
\(\Rightarrow \frac { dy }{ dx } =\frac { x }{ 2 } \)
\(\Rightarrow \left( \frac { dy }{ dx } \right) _{ x=2 }=1\)
The equation of tangent is y = x - 1.

The required Area = Shaded Area of graph
\(\Rightarrow -\left[ \int _{ 2 }^{ 3 }{ \left\{ \left( x-1 \right) -\frac { x }{ 2 } \right\} dx+\int _{ 3 }^{ 6 }{ \left[ \frac { \left( x+3 \right) }{ 3 } -\frac { x }{ 2 } \right] dx } } \right] \)
\(\Rightarrow -\left[ \int _{ 2 }^{ 3 }{ \left( x-1 \right) dx+\frac { 1 }{ 3 } \int _{ 3 }^{ 6 }{ \left( x+3 \right) dx-\frac { 1 }{ 2 } \int _{ 2 }^{ 6 }{ x } dx } } \right] \)
\(\Rightarrow -\left[ \left[ \frac { { x }^{ 2 } }{ 2 } -x \right] _{ 2 }^{ 3 }+\frac { 1 }{ 3 } \left[ \frac { { x }^{ 2 } }{ 2 } +3x \right] _{ 3 }^{ 6 }-\frac { 1 }{ 4 } \left[ { x }^{ 2 } \right] _{ 2 }^{ 6 } \right] \)
\(\Rightarrow \left[ \frac { 9 }{ 2 } -3-2+2 \right] -\frac { 1 }{ 3 } \left[ 18+18-\frac { 9 }{ 2 } -9 \right] +\frac { 1 }{ 4 } \left[ 36-4 \right] \) 
= 1 sq.unit

x² + y² = 32; y = x, point of intersection is y = 4.

Required Area = \(\int _{ 0 }^{ 4 }{ y\quad dy+\int _{ 4 }^{ 4\sqrt { 2 } }{ \sqrt { 32-{ y }^{ 2 }dy } } } \)
\(=\left[ \frac { { y }^{ 2 } }{ 2 } \right] _{ 0 }^{ 4 }+\left[ \frac { y }{ 2 } \sqrt { 32-{ y }^{ 2 } } +16{ sin }^{ -1 }\frac { y }{ 4\sqrt { 2 } } \right] _{ -4 }^{ 4\sqrt { 2 } }\)
\(\Rightarrow =8+\left( 0+16.\frac { \pi }{ 2 } \right) -\left( 8+16.\frac { \pi }{ 4 } \right) =\ 4\pi \) 

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 = \(\int _{ 0 }^{ 1 }{ \sqrt { 3 } x\ dx+\int _{ 1 }^{ 4 }{ 4\ \frac { 1 }{ \sqrt { 3 } } \left( 4-x \right) dx } } \)
\(=\sqrt { 3 } \left[ \frac { { x }^{ 2 } }{ 2 } \right] _{ 0 }^{ 1 }+\frac { 1 }{ \sqrt { 3 } } \left[ 4x-\frac { { x }^{ 2 } }{ 2 } \right] _{ 1 }^{ 4 }\)
\(=\frac { \sqrt { 3 } }{ 2 } +\frac { 1 }{ \sqrt { 3 } } \left[ 16-8-4+\frac { 1 }{ 2 } \right] \)
\(=2\sqrt { 3 } \ sq.units\)