The given plane is passing through the point (1, 1, 1) and parallel to two vectors \((\hat{i}-\hat{j})\) and \((\hat{j}-\hat{k})\)
The corresponding Cartesian equation is of the form 
\(\left|\begin{array}{ccc} x-x_{1} & y-y_{1} & z-z_{1} \\ l_{1} & m_{1} & n_{1} \\ l_{2} & m_{2} & n_{2} \end{array}\right|=0\left\{\begin{array}{c} \left(x_{1}, y_{1}, z_{1}\right)=(1,1,1) \\ \left(l_{1}, m_{1}, n_{1}\right)=(1,-1,0) \\ \left(l_{2}, m_{2}, n_{2}\right)=(0,1,-1) \end{array}\right.\)
\(\text { i.e., }\left|\begin{array}{ccc} x-1 & y-1 & z-1 \\ 1 & -1 & 0 \\ 0 & 1 & -1 \end{array}\right|=0 \text { i.e., } x+y+z-3=0\)
Here \((\left.x_{1}, y_{1}, z_{1}\right)\) = (1, -1, 2)
Distance \( =\left|\frac{a x_{1}+b y_{1}+c z_{1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|=\left|\frac{1-1+2-3}{\sqrt{1+1+1}}\right| \)
\( =\frac{1}{\sqrt{3}} \text { units } \)

The equation of the line passing through (1, 1, -1) and (-1,0, 1) is
\(\frac{x-1}{2}=\frac{y-1}{1}=\frac{z+1}{-2}\)
It meets the xy-plane i.e. z = 0
\(\therefore \frac{x-1}{2}=\frac{y-1}{1}=\frac{1}{-2} \Rightarrow x=0, y=\frac{1}{2}\)
The required point is \(\left(0, \frac{1}{2}, 0\right)\)

\(a\left( \frac { 3 }{ 2 } \right) =2\Rightarrow a\frac { 4 }{ 3 } \)
b² = a²(e² - 1)
\(\Rightarrow b^{2}=\frac{16}{9}\left(\frac{9}{4}-1\right)=\frac{16}{9}\left(\frac{9-4}{4}\right)=\frac{\not 16}{9} \times \frac{5}{4}\)
⇒ \({ b }^{ 2 }=\frac { 4\times 5 }{ 9 } =\frac { 20 }{ 9 } \)
∴ Equation of the hyperbola is \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\)
⇒ \(\frac { { x }^{ 2 } }{ \frac { 16 }{ 9 } } -\frac { { y }^{ 2 } }{ \frac { 20 }{ 9 } } =1\)
⇒ \(\frac { { 9x }^{ 2 } }{ 16 } -\frac { 9{ y }^{ 2 } }{ 20 } =1\)

The given equation is of the hyperbola is
\(\frac{x^{2}}{9}-\frac{y^{2}}{25}=1\)
We know that the point P (x, y) lies outside, on or inside the hyperbola
\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)
According as \(\frac{x_{1}^{2}}{a^{2}}-\frac{y_{1}^{2}}{b^{2}}-1<0,=\text { or }>0\)
According to the given problem,
\( \frac{x_{1}^{2}}{a^{2}}-\frac{y_{1}^{2}}{b^{2}}-1 =\frac{6^{2}}{9}-\frac{(-5)^{2}}{25}-1 \)
\( =\frac{26}{9}-\frac{25}{25}-1 \)
\( =4-1-1 =2>0 \)
Therefore, the point (6, - 5) lies inside the hyperbola
 \(\frac{x^{2}}{9}-\frac{y^{2}}{25}=1\)

Let \(\cos ^{-1} x=\theta \text {, then } \cos \theta=x, \text { where } \theta \in[0, \pi]\)
Therefore \(\tan \left(\cos ^{-1} x\right)=\tan \theta=\frac{\sin \theta}{\cos \theta}\)
\(=\frac{\sqrt{1-\cos ^{2} \theta}}{\cos \theta}=\frac{\sqrt{1-x^{2}}}{x}\)
Hence \(\tan \left(\cos ^{-1} \frac{8}{17}\right)=\frac{\sqrt{1-\left(\frac{8}{17}\right)^{2}}}{\frac{8}{17}}=\frac{15}{8}\)

Let \(\vec{r}=\vec{a}+t \vec{b}\) be the equation of the given line which is not parallel to the given plane whose equation is \(\text { So, } \vec{b} \cdot \vec{n} \neq 0\) .
Let \(\vec{u}\) be the position vector of the meeting point of the line with the plane. Then \(\vec{u}\) satisfies both \(\vec{r}=\vec{a}+t \vec{b}\) and \(\vec{r} \cdot \vec{n}=p\) for some value of t , say t₁. So, We get
\( \vec{u}=\vec{a}+t \vec{b} \) ...(1)
\(\vec{u} \cdot \vec{n}=p\)   ...(2)
Substuting (1) in (2), we get
\( \left(\vec{a}+t_{1} \vec{b}\right) \cdot \vec{n}=p \)
\(\text { or } \quad \vec{a} \cdot \vec{n}+t_{1}(\vec{b} \cdot \vec{n})=p \ \)
\(\text { or } \ \ t_{1}=\frac{p-(\vec{a} \cdot \vec{n})}{\vec{b} \cdot \vec{n}}\)
\(\text { or } \ t_{1}=\frac{p-(\vec{a} \cdot \vec{n})}{\vec{b} \cdot \vec{n}}\)
Substuting (3) in (1), we get
\(\vec{u}=\vec{a}+\left(\frac{p-(\vec{a} \cdot \vec{n})}{\vec{b} \cdot \vec{n}}\right) \vec{b}, \vec{b} \cdot \vec{n} \neq 0\)

Let A (x₁ , y₁ , z₁) be any point on the plane  ax + by + cz + d₂ = 0 , then we have
\(a x_{1}+b y_{1}+c z_{1}+d_{2}=0 \Rightarrow a x_{1}+b y_{1}+c z_{1}=-d_{2}\)
The distance of the plane  ax + by + cz + d₁ = 0 from the point A (x₁ , y₁ , z₁) is given by
\(\delta=\frac{\left|a x_{1}+b y_{1}+c z_{1}+d_{1}\right|}{\sqrt{a^{2}+b^{2}+c^{2}}}=\frac{\left|d_{1}-d_{2}\right|}{\sqrt{a^{2}+b^{2}+c^{2}}}\)
Hence, the distance between two parallel planes ax + by + cz + d₁ = 0 and ax + by + cz + d₂ = 0 is given by \(\delta=\frac{\left|d_{1}-d_{2}\right|}{\sqrt{a^{2}+b^{2}+c^{2}}}\)

The two skew lines \(\vec{r}=\vec{a}+s \vec{b} \text { and } \vec{r}=\vec{c}+t \vec{d}\) are denoted by L₁ and  L₂ respectively.
Let A and C be the points on  L₁ and  L₂ with position vectors \(\vec{a} \text { and } \vec{c}\) respectively.
From the given equations of skew lines, we observe that L₁ is parallel to the vector \(\vec{b}\) and  L₂ is parallel to the vector \(\vec{d}\) So, \(\vec{b} \times \vec{d}\) is perpendicular to the lines L₁ and  L₂ .
Let SD be the line segment perpendicular to both the lines  L₁ and  L₂ . Then the vector \(\overrightarrow{S D} \) is perpendicular to the vectors \(\vec{b}\) and \(\vec{d}\) and therefore it is parallel to the vector \(\vec{b} \times \vec{d}\) .
So, \(\frac{\vec{b} \times \vec{d}}{|\vec{b} \times \vec{d}|}\) is a unit vector in the direction of \(\overrightarrow{S D} \) Then, the shortest distance \(|\overrightarrow{S D}|\) is the absolute value of the projection of \(\overrightarrow{AC} \) and \(\overrightarrow{S D} \). That is
\(\delta=|\overrightarrow{S D}|=\mid \overrightarrow{A C}\) (Unit vector in the direction of \(\overrightarrow{S D})|=|(\vec{c}-\vec{a}) \cdot \frac{\vec{b} \times \vec{d}}{|\vec{b} \times \vec{d}|} \mid\)
\(\delta=\frac{|(\vec{c}-\vec{a}) \cdot(\vec{b} \times \vec{d})|}{|\vec{b} \times \vec{d}|}, \text { where }|\vec{b} \times \vec{d}| \neq 0\)

If \(\vec{a}, \vec{b}, \vec{c}\) are non-coplanar and
\(\left|\begin{array}{lll} x_{1} & y_{1} & z_{1} \\ x_{2} & y_{2} & z_{2} \\ x_{3} & y_{3} & z_{3} \end{array}\right| \neq 0\)
then the three vectors \(\vec{p}=x_{1} \vec{a}+y_{1} \vec{b}+z_{1} \vec{c}, \quad \vec{q}=x_{2} \vec{a}+y_{2} \vec{b}+z_{2} \vec{c}, \text { and }, \vec{r}=x_{3} \vec{a}+y_{3} \vec{b}+z_{3} \vec{c}\) are also non-coplanar.

Let A (x₁, y₁) and B (x₂, y₂) be the two extremities of the diameter AB and P(x, y) be any point on the circle. Then \(\angle A P B=\frac{\pi}{2}\) (angle in a semi-circle).
Therefore, the product of slopes of AP and PB is equal to -1.
That is, \(\left(\frac{\left(y-y_{1}\right)}{\left(x-x_{1}\right)}\right)\left(\frac{\left(y-y_{2}\right)}{\left(x-x_{2}\right)}\right)=-1\) yielding the equation of the required circle as (x − x ₁)(x − x₂ ) + ( y − y₁ )( y − y2 ) = 0

Let the circle be S : x² + y² +2gx+2 fy+c = 0 , … (1)
and the line be L : lx+my+n = 0 . … (2)
Consider S +\(\lambda\) L = 0 . That is x² + y² + 2gx + 2 fy + c +\(\lambda\) (lx + my + n) = 0    ....(3)
Grouping the terms of x, y and constants, we get \(x^{2}+y^{2}+x(2 g+\lambda l)+y(2 f+\lambda m)+c+\lambda n=0\) which is a second degree equation in x and y with coefficients of x² and y² equal and there is no xy term.
If (\(\alpha\), \(\beta\) ) is a point of intersection of S and L satisfying equation (1) and (2), then it satisfies equation (3).
Hence S + \(\lambda\)L = 0 represents the required circle.

\({ cos }^{ -1 }\left( cos\frac { \pi }{ 7 } cos\frac { \pi }{ 17 } -sin\frac { \pi }{ 17 } sin\frac { \pi }{ 17 } \right) .\)
\({ cos }^{ -1 }\left( cos\left( \frac { \pi }{ 7 } +\frac { \pi }{ 17 } \right) \right) \)
\(\left[ \therefore cosA\ cosB-sinA\ sinB=cos(A+B) \right] \)
= \({ cos }^{ -1 }\left( cos\left( \frac { 24\pi }{ 119 } \right) \right) \)    \(\left[ \therefore \frac { 24\pi }{ 119 } \varepsilon \left[ 0,\pi \right] \right] \)
= \(\frac { 24\pi }{ 119 } \) 

Equation of given plane is
\(\vec { r } .\left( 6\hat { i } +3\hat { j } +2\hat { k } \right) =8\)
\(\therefore { \vec { n } }_{ 1 }=6\hat { i } +3\hat { j } +2\hat { k } \)
and the line is \(\vec { r } =\left( 2\hat { i } -\hat { j } +\hat { k } \right) +t\left( \hat { i } +2\hat { j } -2\hat { k } \right) \)
\(\therefore \vec { b } =\hat { i } +2\hat { j } -2\hat { k } \)
Angle between a line and a plane is
\(sin\theta =\frac { \left| \vec { b } .\vec { n } \right| }{ \left| \vec { b } \right| \left| \vec { n } \right| } =\cfrac { \left| 6(1)+3(2)+2(-2) \right| }{ \sqrt { { 1 }^{ 2 }+{ 2 }^{ 2 }+\left( -2 \right) ^{ 2 }.\sqrt { { 6 }^{ 2 }+{ 3 }^{ 2 }+{ 2 }^{ 2 } } } } \)
= \(\frac { 6+6-4 }{ \sqrt { 9 } .\sqrt { 36+9+4 } } =\frac { 8 }{ \sqrt { 9 } .\sqrt { 49 } } =\frac { 8 }{ 3\left( 7 \right) } =\frac { 8 }{ 21 } \)
\(\therefore \theta ={ sin }^{ -1 }\left( \cfrac { 8 }{ 21 } \right) \)

Given \(\vec { a } \) = \(2\hat { i } +6\hat { j } +3\hat { k } \)
\(\vec { n } \) = \(\hat { i } +3\hat { j } +5\hat { k } \)
Vector form of the evaluation of the plane passing through one point (\(\vec { a } \)) and normal to a vector (\(\vec { n } \)) is
\(\vec { r } .\vec { n } =\vec { a } .\vec { n }\)
\(=\vec { r } .(\hat { i } +3\hat { j } +5\hat { k } )=(2\hat { i } +6\hat { j } +3\hat { k } ).(\hat { i } +3\hat { j } +5\hat { k } )\)
= 2 + 18 + 15
\(\Rightarrow \vec { r } .(\hat { i } +3\hat { j } +5\hat { k } )=35\)
Its Cartesian equation will be
\(a(x-{ x }_{ 1 })+b(y-{ y }_{ 1 })+c(z-{ z }_{ 1 })=0\)
1(x-2)+3(y-6)+5(z-3) = 0
[\(\because ({ x }_{ 1 },{ y }_{ 1 },{ z }_{ 1 }\) is (2, 6, 3) and a, b, c = 1, 3, 5]
\(\Rightarrow\) x-2 + 3y-18 + 5z-15 = 0
\(\Rightarrow\) x + 3y + 5z - 35 = 0
\(\Rightarrow\) x + 3y + 5z = 35

The equation of the ellipse is \(\frac { { \left( x-11 \right) }^{ 2 } }{ 484 } +\frac { { y }^{ 2 } }{ 64 } =1\). The origin of the sound wave and the kidney stone of patient should be at the foci in order to crush the stones.
a² = 484 and b² = 64
c² = a² -b²
= 484-64
= 420
c \(\simeq \) 20.5
Therefore the patient’s kidney stone should be placed 20.5 cm from the centre of the ellipse.

Equation of the parabola is y = \(\frac { 1 }{ 32 } \)x²
That is x² = 32y ; the vertex is (0, 0)
= 4 (8)y
\(\Rightarrow a=8\)
So the heating tube needs to be placed at focus (0, a) 
Hence the heating tube needs to be placed 8 units above the vertex of the parabola.

\({ tan }^{ -1 }\left( \frac { x }{ y } \right) -{ { tan }^{ -1 }\left( \frac { x-y }{ x+y } \right) }\)
= \(tan^{ -1 }\left( \frac { \frac { x }{ y } -\frac { x-y }{ x+y } }{ 1+\frac { x }{ y } \left( \frac { x-y }{ x+y } \right) } \right) \)
\(\left[ \because { tan }^{ -1 }x-{ tan }^{ -1 }y={ tan }^{ -1 }\left( \frac { x-y }{ 1+xy } \right) \right] \)
= \({ tan }^{ -1 }\left( \frac { \frac { x(x+y)-y(x-y) }{ y(x+y) } }{ \frac { y(x+y)+x(x-y) }{ y(x+y) } } \right) \)

= tan^-1(1)
= \(\frac { \pi }{ 4 } \)

Let P(x, y) be any point on the hyperbola
\(\frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\)

Then by definition, SP = ePM and S'P = ePM'
SP = ePM ⇒ SP = e(NK)
= e(CN - CK)
= \(e\left( x-\frac { a }{ e } \right) \)  = ex - a
= and S'P = ePM' ⇒ S'P = e(NK')
⇒ e(CN + CK') = \(e\left( x+\frac { a }{ e } \right) \) = ex + a
∴ S'P -SP = (ex + a) - (ex - a)
ex + a - ex + a = 2a (constant)
= length of the transverse axis.

 Passing through (5, -2) length of the transverse axis is a long x-axis and of length 8 units.
2a = 8 ⇒ a = 4
Since the transverse axis is along x-axis, centre is (0, 0)
Equation of the hyperbola is
\(\frac { { x }^{ 2 } }{ 16 } -\frac { { y }^{ 2 } }{ b^{ 2 } } =1\)
Since (5, -2)passes through the parabola,
\(\frac { 25 }{ 16 } -\frac { 4 }{ { b }^{ 2 } } \Rightarrow \frac { 4 }{ { b }^{ 2 } } =\frac { 25 }{ 16 } -1=\frac { 25-16 }{ 16 } =\frac { 9 }{ 16 } \)
∴ \({ b }^{ 2 }=\frac { 16\times 4 }{ 9 } =\frac { 64 }{ 9 } \)
∴ Equation of the hyperbola is
\(\frac { { x }^{ 2 } }{ 16 } -\frac { { y }^{ 2 } }{ \frac { 64 }{ 9 } } =1\Rightarrow \frac { { x }^{ 2 } }{ 16 } -\frac { 9{ y }^{ 2 } }{ 64 } =1\)

Let ABC be an isosceles triangle with AB = AC and let AD is the median 
D is mid-point of BC.
\( \overrightarrow{A D}=\frac{1}{2}(\overrightarrow{A B}+\overrightarrow{A C}) \)
\( \overrightarrow{B C}=\overrightarrow{B A}+\overrightarrow{A C} \)
\( \overline{D A} \cdot \overrightarrow{D B}=-\overrightarrow{A D} \cdot\left(\frac{-1}{2} \overrightarrow{C B}\right) \)
\(=-\overrightarrow{A D} \cdot\left(\frac{1}{2} \overrightarrow{B C}\right) \)
\(=\frac{1}{2} \overrightarrow{A D} \cdot \overrightarrow{B C} \)
\(=\frac{1}{4}(\overrightarrow{A B}+\overrightarrow{A C}) \cdot(\overrightarrow{B A}+\overrightarrow{A C}) \)
\(=\frac{1}{4}(\overrightarrow{A B}+\overrightarrow{A C}) \cdot(\overrightarrow{A C}-\overrightarrow{A B}) \)
\(=\frac{1}{4}[(\overrightarrow{A C} \cdot \overrightarrow{A C})-(\overrightarrow{A B} \cdot \overrightarrow{A B})] \)
\(=\frac{1}{4}\left(A C^{2}-A B^{2}\right) \)
\(=\frac{1}{4}(0)=0 \)
\(\overrightarrow{D A} \cdot \overrightarrow{D B}=0\)
\(\overrightarrow{D A} \perp \overrightarrow{D B}\)

Parabola is open left and axis of symmetry as x-axis and vertex (0, 0)  
Then the equation of the required parabola is
(y - 0)² = -4\(\sqrt { 2 } \) (x - 0)
y² = -4\(\sqrt { 2 } \) x

Given equation of the circle is
3x² + (3 - p)xy + qy² - 2px = 8pq
For the circle, co-efficient of xy = 0
⇒ 3-p = 0 ⇒ p = 3
Also, co-efficient of x² = co-efficient of y²
⇒ 3 = q
∴ Equation of the circle is
3x² + 3y² - 6x = 8(3)(3)
3x² + 3y² - 6x - 72 = 0
Dividing by 3, we get
x² + y² - 2x - 24 = 0
Here 2g = - 2⇒ g = - 1
f = 0 and c = - 24
Centre is (-g, -f) (1, 0)
and r = \(\sqrt { { g }^{ 2 }+{ f }^{ 2 }-c } \)
= \(\sqrt { { (-1 })^{ 2 }+0+24 } \)
= \(\sqrt { 25 } \) = 5 units.

Since the circles touch both the axes. Its equation will be
(x + a)² + (y + a)² = a² ...............(1)
It passes through (-4, -2)
∴ (-4 + a)² + (-2 + a)² = a²
\(16+\not a^{2}+8 a+4+a^{2}+4 a=\not a^{2}\)
⇒ a² + 12a + 20 = 0
⇒ (a + 10)(a + 2) = 0
a = -10 or -2

Case (i):
When a = -10, (1) becomes
(x + 10)² + (y + 10)2 = 10²
\(\Rightarrow x^{2}+\not 100+20 x+y^{2}+\not 100+20 y=160\)
⇒  x² + y²+ 20x + 20y + 100 = 0
Case (ii):
When a = -2, (1) becomes
⇒ (x + 2)² + (y + 2)² = 2²
\(x^{2}+4 x+4+y^{2}+4 y+\not 4 = \not 4\)
x² + y²+ 4x + 4y + 4 = 0
Hence, equation of the circles are
x² + y²+ 4x + 4y + 4 = 0
or  x² + y²+ 20x + 20y + 100 = 0

Given centre is (2, -1) and passing through the point (3, 6)
∴ r = distance between (2, -1) and (3, 6)
= \(\sqrt { { (2-3) }^{ 2 }+(-1{ -6) }^{ 2 } } \)
= \(\sqrt { { (-1) }^{ 2 }+({ -7) }^{ 2 } } \)
= \(\sqrt { 1+49 } =\sqrt { 50 } \)
∴ Equation of the circle is
(x - h)² + (y - k)² = r²
(x−2)²+(y+1)² = \({ (\sqrt { 50 } })^{ 2 }\)
(x−2)²+(y+1)² = 50