Given lines are
\(\frac { x-1 }{ 2 } =\frac { y+1 }{ 3 } =\frac { z-1 }{ 4 } \)
\(\vec { a } =\hat { i } -\hat { j } +\hat { k } \vec { b } \quad and\quad \vec { b } =2\hat { i } +3\hat { j } +4\hat { k } \)
\(\frac { x-3 }{ 1 } =\frac { y-m }{ 2 } =\frac{z-0}{1}\)\(\Rightarrow \vec { c } =3\hat { i } +m\hat { j } \)
\(\vec { d } =\hat { i } +2\hat { j } +\hat { k } \quad \)
\(\vec { c } -\vec { a } =(3\hat { i } +m\hat { j } )-(\hat { i } -\hat { j } +\hat { k } )\)
\(=2\hat { i } +(m,+1)\hat { j } -\hat { k } \)
\(\vec { b } \times \vec { d } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 2 & 3 & 4 \\ 1 & 2 & 1 \end{matrix} \right| =\hat { i } (3-8)-\hat { j } (2-4)+\hat { k } (4-3)\)
\(=-5\hat { i } +2\hat { j } +\hat { k } \)
Since the lines intersect at a point, \((\vec { c } -\vec { a } ).(\vec { b } \times \vec { d } )=0,\)
\(\Rightarrow (2\hat { i } +(m+1)\hat { j } -\hat { k } ).(-5\hat { i } +2\hat { j } +\hat { k } )=0\)
\(\Rightarrow -10+2(m+1)-1=0\Rightarrow -10+2m+2-1=0\)
\(\Rightarrow 2m-9=0\Rightarrow 2m=9\Rightarrow m=\frac { 9 }{ 2 } \)
\(\therefore m=\frac { 9 }{ 2 } \)

Given lines are \(\frac { x-5 }{ 5m+2 } =\frac { 2-y }{ 5 } =\frac { 1-z }{ -1 } \)
⇒ \(\frac { x-5 }{ 5m+2 } =\frac { y-2 }{ -5 } =\frac { z-1 }{ 1 } \)
∴ \(\vec { b } =(5m+2)\hat { i } -5\hat { j } +\hat { k } \) ...(1)
and x = \(\frac { 2y+1 }{ 4m } =\frac { 1-z }{ -3 } \)
⇒ \(\frac { x }{ 1 } =\frac { y+\frac { 1 }{ z } }{ 2m } =\frac { z-1 }{ 3 } \)
∴ \(\vec { d } =\hat { i } +2m\hat { j } +3\hat { k } \) ...(2)
since \(\vec { b } \bot \vec { d } \Rightarrow \vec { b } .\vec { d } \)= 0
⇒ \((\hat { i } +2m\hat { j } +3\hat { k } ).((5m+2)\hat { i } -5\hat { j } +\hat { k } )=0\)
⇒ (5m+2)1+2m(-5)+3(1) = 0
⇒ 5m+2-10m+3 = 0
⇒ 5-5m = 0
⇒ 5 = 5m
∴ m = 1

Let \(\frac { \sqrt { 3 } }{ 2 } +\frac { i }{ 2 } \) = r(cos θ + i sin θ)
r = \(\sqrt { \left( \frac { \sqrt { 3 } }{ 2 } \right) ^{ 2 }+\left( \frac { 1 }{ 2 } \right) ^{ 2 } } =\sqrt { \frac { 3 }{ 4 } +\frac { 1 }{ 4 } } =\sqrt { \frac { 4 }{ 4 } } \)=1
α = \(tan^{ -1 }\left| \frac { y }{ x } \right| =tan^{ -1 }\left| \frac { \frac { 1 }{ 2 } }{ \frac { \sqrt { 3 } }{ 2 } } \right| =tan^{ -1 }\left( \frac { 1 }{ \sqrt { 3 } } \right) =\frac { \pi }{ 6 } \)
Since \(\frac { \sqrt { 3 } }{ 2 } +\frac { i }{ 2 } \) lies is the I quadrant, θ = α
∴ \(\frac { \sqrt { 3 } }{ 2 } +\frac { i }{ 2 } =1\left( cos\frac { \pi }{ 6 } +isin\frac { \pi }{ 6 } \right) \)
∴ \(\left( \frac { \sqrt { 3 } }{ 2 } +\frac { i }{ 2 } \right) ^{ 5 }=1^{ 5 }\left( cos\frac { \pi }{ 6 } +isin\frac { \pi }{ 6 } \right) ^{ 5 }\)
= \(cos\frac { 5\pi }{ 6 } +isin\frac { 5\pi }{ 6 } \)  ....(1) [De moivres theorem]
Similarly \(\frac { \sqrt { 3 } }{ 2 } +\frac { i }{ 2 } =1\left( cos\frac { \pi }{ 6 } -isin\frac { \pi }{ 6 } \right) \)
∴ \(\left( \frac { \sqrt { 3 } }{ 2 } -\frac { i }{ 2 } \right) ^{ 5 }=1^{ 5 }\left[ cos\frac { \pi }{ 6 } +isin\frac { \pi }{ 6 } \right] ^{ 5 }\)
= \(cos\frac { 5\pi }{ 6 } -isin\frac { 5\pi }{ 6 } \) ....(2)
Adding (1) and (2) we get,
\(\left( \frac { \sqrt { 3 } }{ 2 } +\frac { i }{ 2 } \right) ^{ 5 }+\left( \frac { \sqrt { 3 } }{ 2 } -\frac { i }{ 2 } \right) ^{ 5 }\)

= \(2cos\frac { 5\pi }{ 6 } =2cos\left( \pi -\frac { \pi }{ 6 } \right) \)
= \(-2cos\ \frac { \pi }{ 6 } \) [∵ \(\frac { 5\pi }{ 6 } \) lies in the II quard]
= \(-2\left( \frac { \sqrt { 3 } }{ 2 } \right) =-\sqrt { 3 } \).

Let O be the centre of the semi-circle and AA¹ be the diameter.
Let P be any point on the circumference of the semi circle.
Taking O as the origin, let the position vectors of A and P be a and \(\vec { r } \) respectively.
Let us prove that \(\angle A P B=90^{\circ}\)
W.K.T OA = OB = OP ( because of radius)
\( \overrightarrow{P A} =\overrightarrow{P O}+\overrightarrow{O A} \)
\(\overrightarrow{P B} =\overrightarrow{P O}+\overrightarrow{O B} \)
\( =\overrightarrow{P O}-\overrightarrow{O A} \)
\(\overrightarrow{P A} \cdot \overrightarrow{P B} =(\overrightarrow{P O}+\overrightarrow{O A})(\overrightarrow{P O}-\overrightarrow{O A}) \)
\( =\overrightarrow{P O}^{2}-\overrightarrow{O A}^{2}=0 \)
\( \overrightarrow{P A} \perp \overrightarrow{P B} \)
\( \Rightarrow \ \angle A P B=90^{\circ} \). Hence proved.

\(\left| z+3+4i \right| \le \left| z \right| +\left| 3+4i \right| =2+5=7\)
\(\left| z+3+4i \right| \le 7\) .............. (1)
\(\left| z+3+4i \right| \ge \left| \left| z \right| -\left| 3+4i \right| \right| =\left| 2-5 \right| =3\)
\(\left| z+3+4i \right| \ge 3\) ............ (2)
From (1) and (2) we get, \(3\le \left| z+3+4i \right| \le 7\) 

The condition for the line y = mx + c to be a tangent to the circle x² + y² = a² is c² = a²(1 + m²) from  
Then \(c=\pm \sqrt { 9\left( 1+16 \right) } \)
\(c=\pm 3\sqrt { 17 } \)

As the circle touches both the axes, the distance of the centre from both the axes is 3 units, centre can be (±3, ±3) and hence there are four circles with radius 3, and the required equations of the four circles are
x² + y² ± 6x ± 6y + 9 = 0.

We know that the formula for the distance between two parallel ax + by + cz + d₁ = 0 and ax + by + cz + d₂ = 0 is \(\delta =\frac { |{ d }_{ 1 }-{ d }_{ 2 }| }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } } } \). Rewrite the second equation as x + 2y - 2z + \(\frac { 5 }{ 2 } \) = 0. 
Comparing the given equations with the general equations, we get a = 1, b = 2, c = -2, d₁ = 1, d₂ = \(\frac { 5 }{ 2 } \). Substituting these values in the formula, we get the distance
\(\delta =\frac { |{ d }_{ 1 }-{ d }_{ 2 }| }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } } } =\frac { |1-\frac { 5 }{ 2 } | }{ \sqrt { { 1 }^{ 2 }+{ 2 }^{ 2 }+(-2^{ 2 }) } } =\frac { 1 }{ 2 } \) units.

Let \(\vec { a } ={ a }_{ 1 }\hat { i } +{ a }_{ 2 }\hat { j } +{ a }_{ 3 }\hat { k } \)
∴ LHS = \(\hat { i } \times (\vec { a } \times \hat { i } )+\hat { j } \times (\vec { a } \times \hat { j } )+\hat { k } \times \vec { a } \times \hat { k }\)
\((\hat { i }. \hat { i } )\vec { a } -(\hat { i } .\hat { a } )\hat { i } +(\hat { j } .\hat { j } )\vec { a } -(\hat { j } .\vec { a } )\hat { j } +(\hat { k } .\hat { k } )\vec { a } -(\hat { k } .\vec { a } )\hat { k } \)
\([\because \vec { a } \times (\vec { b } \times \vec { c } )=(\vec { a } .\vec { c } )\vec { b } -(\vec { a } .\vec { b } )\vec { c } ]\)
\(1.\vec { a } -{ a }_{ 1 }\hat { i } +1.\vec { a } -{ a }_{ 2 }\hat { j } +1.\vec { a } -{ a }_{ 3 }\hat { k } ]\)
\([\because \hat { i } .\hat { i } =\hat { j } .\hat { j } =\hat { k } .\hat { k } =1\)and
\(\hat { i } \vec { a } =\hat { i } ({ a }_{ 1 }\hat { i } +{ a }_{ 2 }\hat { j } +{ a }_{ 3 }\hat { k } )={ a }_{ 1 }\hat { j } .\vec { a } ={ a }_{ 2 }\quad \hat { k } .\vec { a } ={ a }_{ 3 }\)
\(3\vec { a } -({ a }_{ 1 }\hat { i } +{ a }_{ 2 }\hat { j } +{ a }_{ 3 }\hat { k } )\)
= \(3\vec { a } -\vec { a } =2\vec { a } \)
= RHS .
∴ LHS = RHS. Hence proved

Since the given three vectors are coplanar, we have \(\left| \begin{matrix} 2 & -1 & 3 \\ 3 & 2 & 1 \\ 1 & m & 4 \end{matrix} \right| \) = 0 ⇒ m = -3

We compute \(\left| 6-8i \right| =\sqrt { { 6 }^{ 2 }+\left( -8 \right) ^{ 2 } } =10\)
and applying the formula for square root, we get
\(\sqrt { 6-8i } =\pm \left( \sqrt { \frac { 10+6 }{ 2 } } -i\sqrt { \frac { 10-6 }{ 2 } } \right) \) (\(\therefore\) b is negative\( \frac{b}{|b|}=-1 \))
= \(\pm \left( \sqrt { 8 } +i\sqrt { 2 } \right) \)
= \(\pm \left( 2\sqrt { 2 } -i\sqrt { 2 } \right) \)

Given that 2a = 36.18, 2b = 9.12 , we get
e = \(\sqrt { 1-\frac { { b }^{ 2 } }{ { a }^{ 2 } } } =\frac { \sqrt { { a }^{ 2 }-{ b }^{ 2 } } }{ a } \) = \(\frac { \sqrt { { \left( \frac { 36.18 }{ 2 } \right) }^{ 2 }{ \left( \frac { 9.12 }{ 2 } \right) }^{ 2 } } }{ \frac { 36.18 }{ 2 } } \)
\(\frac { \sqrt { { \left( 18.09 \right) }^{ 2 }-{ \left( 4.56 \right) }^{ 2 } } }{ \left( 8.09 \right) } \approx0.97\)

Given ends of diameter are (3, 4)(2, -7)
∴ Equation of the circle is
(x - x₁)(x - x₂) + (y - y₁)(y - y₂) = 0
⇒ (x - 3)(x - 2) + (y - 4)(y + 7) = 0
⇒ x² - 2x - 3x + 6 + y² + 7y - 4y - 28 = 0
⇒ x² + y² − 5x + 3y − 22 = 0

\(\overline { (2-i)^{ 12 }+(2+i)^{ 12 } } =\overline { (2-i)^{ 12 } } +\overline { (2+i)^{ 12 } } \)
\(\left( 2+i\sqrt { 3 } \right) ^{ 10 }-\left( 2-i\sqrt { 3 } \right) ^{ 10 }\)
Now \(\overline { z } \) = \(\overline { (2+\sqrt { 3 } )^{ 10 }-(2-i\sqrt { 3 } )^{ 10 } } \)
\(\overline { z } \) = \(\overline { (2+i\sqrt { 3 } )^{ 10 } } -(2-i\sqrt { 3 } )^{ 10 }\)
[∵ \(\overline { { z }_{ 1 }-{ z }_{ 2 } } =\overline { { z }_{ 1 } } -\overline { { z }_{ 2 } } \)]
= \((2-i\sqrt { 3 } )^{ 10 }-(2+i\sqrt { 3 } )^{ 10 }\)
= -\(\left[ (2-i\sqrt { 3 } )^{ 10 }-(2+i\sqrt { 3 } )^{ 10 } \right] \)
∴ \(\overline { z } \) = -\(\ { z } \) ⇒ z is purely imaginary
Hence \((2-i\sqrt { 3 } )^{ 10 }-(2+i\sqrt { 3 } )^{ 10 }\)is purely imaginary

Comparing the two given lines with
\(\vec { r } =\vec { a } +t\vec { b } ,\vec { r } =\vec { c } +s\vec { d } \)
we have, \(\vec { a } =-\hat { i } -3\hat { j } -5\hat { k } ,\vec { b } =3\hat { i } +5\hat { j } +7\hat { k } ,\vec { c } =2\hat { i } +4\hat { j } +6\hat { k } \)  and \(\vec { d } =\hat { i } +4\hat { j } +7\hat { k } \)
We know that the two given lines are coplar, if \((\vec { c } -\vec { a } ).(\vec { b } \times \vec { d } )\) = 0
Here, \(\vec { b } \times \vec { d } \left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 3 & 5 & 7 \\ 1 & 4 & 7 \end{matrix} \right| =7\hat { i } -14\hat { j } +7\hat { k } \)  and  \(\vec { c } -\vec { a } =3\hat { i } +7\hat { j } +11\hat { k } \)
Then, \((\vec { c } -\vec { a } ).(\vec { b } \times \vec { d } )=(3\hat { i } +7\hat { j } +11\hat { k } )(7\hat { i } -14\hat { j } +7\hat { k } )=0\)
Therefore the two given lines are coplanar. Then we find the non parametric form of vector equation of the plane containing the two given coplanar lines. We know that the plane containing the two given coplanar lines is
\((\vec { r } -\vec { a } ).(\vec { b } \times \vec { d } )\)= 0
which implies that \((\vec { r } -(-\hat { i } -3\hat { j } -5\hat { k } )).(7\hat { i } -14\hat { j } +7\hat { k } )\) = 0. 
Thus, the required non-parametric vector equation of the plane containing the two given coplanar lines is
\(\vec { r } .(\hat { i } -2\hat { j } +\hat { k } )\) = 0.

As per the given information, we can take the parabola as open downward.

∴ Its equation is x² = -4ay ...(1)
Let P be a point on the flow paths, 2.5 m below the line of the pipe and 3m beyond the vertical line through the end of the pipe.

∴ Pis (3, -2.5)
∴ (1) becomes 3² = -4a(-2.5)
⇒  \(\frac { -9 }{ 2.5 } \) = 4a
∴ (1) becomes, x₁² = \(\frac { -9 }{ 2.5 } \)y .....(2)
Let x₁ be the distance between the bottom of the verticallineon the ground from the pipeendand the point on which this water touches the ground.
But the height of the pipe from the ground is 7.5 m.
∴ (x₁, -7.5) lies on ....(2)
⇒ x₁² = \(\frac { -9 }{ 2.5 } \)(-7.5)
⇒ x₁ = \(\sqrt { 9\times 3 } =3\sqrt { 3 } \) m
The water strikes the ground \(3\sqrt { 3 } \) m beyond the vertical line.

Let the of the parbola be x² = 4ay (1)

Since (30, 16) is a point on (1),
we get 30² = 4 \(\times\) a \(\times\) 16
⇒ a = \(\frac { 30\times 30 }{ 4\times 16 } =\frac { 225 }{ 16 } \)
∴ becomes, x² = \({ x }^{ 2 }=\frac { 4\times 225 }{ 16 } y=\frac { 225 }{ 4 } y\)
Let AC = h m and BD = lm
∴ A(6, h) is a point on the parabola [∵ OD = 6]
∴ \({ 6 }^{ 2 }=\frac { 225 }{ 4 } \times h\)
⇒ \(h=\frac { 36\times 4 }{ 225 } \Rightarrow h=0.52\)
∴ AD = 3 + h = 3 + 0.52 = 3.52 m
Also (12, 1) is a point on the parabola
[∵ ON = 6 + 6 = 12]
∴ \({ 12 }^{ 2 }=\frac { 225 }{ 4 } \times l\)
⇒ l = \(\frac { 12\times 12\times 4 }{ 225 } =\frac { 576 }{ 225 } =2.08\) = 5.08 m
Hence the length of first two vertical cables are 3.52 m and 5.08 m.

Let \(\hat { a } =\vec { OA } \) and \(\hat { b } =\vec { OB } \) be the unit vectors and which make angles α,β respectively with positive x-axis
Draw AL and BM 丄 to x-axis
Then \(|\vec { OL } |=|\vec { OA } |cos\alpha \Rightarrow \vec { OL } =\vec { |OL| } \hat { i } =cos\alpha \hat { i } \)
\(|\vec { LA } |=|\vec { OB } |\) sin α
⇒ \(\vec { LA } =|\vec { OB } |\hat { j } =sin\alpha (-\hat { j } )=-sin\alpha \hat { j } \)
[\(\vec { LA } \) is in the opp direction of y axis]
\(\hat { a } =\vec { OA } =\vec { OL } +\vec { LA } =cos\alpha \hat { i } -sin\alpha \check { j } \) ..(1)
Similarly \(\hat { b } =\vec { OB } =\vec { OM } +\vec { MB } =cos\beta \hat { i } +sin\beta \hat { j } \) ...(2)
Now \(\hat { a } \times \hat { b } =|\hat { a } ||\hat { b } |sin(\alpha +\beta )\hat { k } =sin(\alpha +\beta )\hat { k } \) ....(3)
[\(|\hat { a } |=|\hat { b } |\) = 1]
Also \(\hat { a } \times \hat { b } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ cos\alpha & -sin\alpha & 0 \\ cos\beta & cos\beta & 0 \end{matrix} \right| \)
= \(\hat { i } (0)-\hat { j } (0)+\hat { k } \)(cosα sinβ + sinα cosβ)
= (sin α cos β + cos α sin β)\(\hat { k } \) .(4)
using (3) and (4), sin(α+β) = sin α cos β + cos α sin β

Given z = x + iy
Im \(\left( \frac { 2z+1 }{ iz+1 } \right) \)= 0
⇒ Im\(\left( \frac { 2(x+iy)+1 }{ i(x+iy)+1 } \right) \)= 0
⇒ Im\(\left( \frac { (2x+1)+2iy }{ ix+i^{ 2 }y+1 } \right) \)
⇒ Im\(\left( \frac { (2x+1)+2iy }{ ix-y+1 } \right) \)
\(\left( \frac { (2x+1)+iy }{ (1-y)+ix } \right) \)
Multiply and divide by the conjugate of the denominator
We get Im\(\left( \frac { (2x+1)+2iy }{ (1-y)+ix } \times \frac { (1-y)-ix }{ (1-y)-ix } \right) \)=0
⇒ Im\(\left( \frac { (2x+1)+2iy\times (1-y)-ix }{ (1-y)^{ 2 }+{ x }^{ 2 } } \right) \)
Choosing the imaginably part we get,
\(\frac { (2x+1)(-x)+2y(1-y) }{ (1-y)^{ 2 }+{ x }^{ 2 } } \)
⇒ (2x+1)-x+2y(1-y) = 0
⇒ -2x²-x+2y-2y² = 0
⇒ 2x²+2y²+x-2y = 0
Hence, locus of z is 2x²+2y²+x-2y = 0

Since,\(\left| { z }_{ 1 } \right| =\left| { z }_{ 2 } \right| =\left| { z }_{ 3 } \right| =1\)
\(\left| { z }_{ 1 } \right| ^{ 2 }=1\Rightarrow { z }\bar { { z }_{ 1 } } =1,\left| { z }_{ 2 } \right| ^{ 2 }=1\Rightarrow { z }_{ 2 }\bar { { z }_{ 2 } } =1\ \)
Therefore, \(\bar { { z }_{ 1 } } =\frac { 1 }{ { z }_{ 1 } } ,\bar { { z }_{ 2 } } =\frac { 1 }{ { z }_{ 3 } } \) and hence
\(\left| \frac { 1 }{ { z }_{ 1 } } +\frac { 1 }{ { z }_{ 2 } } +\frac { 1 }{ { z }_{ 3 } } \right| =\left| \bar { { z }_{ 1 } } +\bar { { z }_{ 2 } } +{ \bar { z } }_{ 3 } \right| \)
= \(\left| \overline { { z }_{ 1 }+\left| { z }_{ 2 }+{ z }_{ 3 } \right| } \right| ={ { z }_{ 1 }+{ z }_{ 2 }+{ z }_{ 3 } }=1\)

Completing the square on x and y of 4x²+36y²+40x−288y+532 = 0,
4(x² + 10x + 25 − 25) + 36(y² − 8y + 16 − 16) + 532 = 0 , gives
4(x² + 10x + 25) + 36(y² − 8y + 16) = −532 + 100 + 576
4(x + 5)² + 36(y − 4)² = 144.
Dividing both sides by 144, the equation reduces to \(\frac { { \left( x+5 \right) }^{ 2 } }{ 36 } \frac { { \left( y-4 \right) }^{ 2 } }{ 4 } =1\)
This is an ellipse with centre (-5, 4), major axis is parallel to x-axis, length of major axis is 12 and length of minor axis is 4. Vertices are (1, 4) and (-11, 4).
Now, c² = a²−b² = 36 − 4 = 32
and c = ±4 \(\sqrt { 2 } \)
Then the foci are (−5 − 4\(\sqrt { 2 } \), 4) and (−5 + 4\(\sqrt { 2 } \) , 4).
Length of the major axis = 2a = 12 units and
the length of the minor axis = 2b = 4 units.