2x = 3y =  −z \(\Rightarrow \frac{x}{3}=\frac{y}{2}=\frac{-z}{6}\) (Dividing by 6 all)
 \(\Rightarrow \frac{x-0}{3}=\frac{y-0}{2}= \frac{z-0}{-6}\)....(1)
6x = -y = -4z \(\Rightarrow \frac{x}{2}=\frac{-y}{12}=\frac{-z}{3} \) (Dividing by 6 all)
\(\Rightarrow \frac{x-0}{2}=\frac{y-0}{-12}=\frac{z-0}{-3}\) ....(2)
From (1) & (2), we get
\(\vec b = 3\vec i+2\vec j- 6\vec k\)and \( \vec d = 2\vec i-12\vec j- 3\vec k\)
Angle between lines (1) and (2) = Angle between \(\vec b\ and\ \vec d\)
Acute angle between lines cos 0 = \(\frac{|\vec b . \vec d|}{|\vec b||\vec d|} \)
\(\vec b . \vec d \)= (\(\vec b = 3\vec i+2\vec j- 6\vec k\)). (\( 2\vec i-12\vec j- 3\vec k\))
6-24+18 = 0
\( \cos \theta=0 \)
\(\theta=\frac{\pi}{2} \text { or } 90^{\circ}\)

Given lines are \(\frac { x+4 }{ 3 } =\frac { y-7 }{ 4 } =\frac { z+5 }{ 5 } \)
⇒ \(\vec { b } =3\hat { i } +4\hat { j } +5\hat { k } \)
and \(\vec { r } =4\hat { k } +t(2\hat { i } +\hat { j } +\hat { k } )\)
⇒ \(\vec { d } =2\hat { i } +\hat { j } +\hat { k } \)
\( |\vec{b}| =\sqrt{9+16+25}=\sqrt{50} \\ =5 \sqrt{2} \)
∴ cos θ =\(\frac { \vec { b } .\vec { d } }{ |\vec { b } ||\vec { d } | } \)=\(\frac { (3\hat { i } +4\hat { j } +5\hat { k } ).(2\hat { i } +\hat { j } +\hat { k } ) }{ \sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 }+{ 5 }^{ 2 } } .\sqrt { { 2 }^{ 2 }+{ 1 }^{ 2 }+{ 1 }^{ 2 } } } \)
 \( = \frac{15}{5 \sqrt{2} \times \sqrt{6}}=\frac{3}{\sqrt{2} \cdot \sqrt{2} \sqrt{3}}=\frac{3}{2 \sqrt{3}}=\frac{\sqrt{3}}{2} \\ \theta=\frac{\pi}{6} \)

Equation of the circle is
2x² + 2y² - 6x + 4y + 2 = 0
Dividing by 2, we get
x² + y² - 3x + 2y + 1 = 0
Here 2g = -3 ⇒ g = \(\frac { -3 }{ 2 } \)
2f = 2 ⇒ f = 1
and c = 1
∴ Centre is (-g, -f) = \(\left( \frac { 3 }{ 2 } ,-1 \right) \)
and r = \(\sqrt { { g }^{ 2 }+{ f }^{ 2 }-c } \) = \(\sqrt { { \left( \frac { 3 }{ 2 } \right) }^{ 2 }+{ 1 }^{ 2 }-1 } \)
= \(\sqrt { \frac { 9 }{ 4 } } =\frac { 3 }{ 2 } \) units.

Equation of the circle is x² + y² - x + 2y - 3 = 0
Here 2g = -1 ⇒ g = \(\frac { -1 }{ 2 } \)
2f = 2 ⇒ f = 1 and c = -3
Centre is (-g, -f) = \(\left( \frac { 1 }{ 2 } ,-1 \right) \)
and r = \(\sqrt { { g }^{ 2 }+{ f }^{ 2 }-c } \)
= \(\sqrt { \frac { 1 }{ 4 } +1+3 } \)
= \(\sqrt { \frac { 1 }{ 4 } +4 } =\sqrt { \frac { 1+16 }{ 2 } } \)
r = \(\sqrt { \frac { 17 }{ 2 } } \) units.

Equation of the circle is
x² + y² + 6x - 4y + 4 = 0.
Here 2g = 6 ⇒ g = 3
2f  = -4 ⇒ f = -2 and c = 4
Centre is (-g, -f) ⇒ (-3, 2)
r = \(\sqrt { { g }^{ 2 }+{ f }^{ 2 }-c } \) = \(\sqrt { { 3 }^{ 2 }+(-2){ }^{ 2 }-{ 4 } } \)
= \(\sqrt { 9+4-4 } \)
= \(\sqrt { 9 } \)
= 3 unit

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.

Comparing the given equation of the plane with \(\vec { r } .\vec { n } \) = p, we have \(\vec { n } =6\hat { i } -3\hat { j } +2\hat { k } \).
We know that the perpendicular distance from the given point with position vector u to the plane \(\vec { r } .\vec { n } \)= p is given by \(\delta =\frac { |\vec { u } .\vec { n } -p| }{ |\vec { n } | } \). Therefore, substi \(\vec { u } \)= (2, 5, -3) = \(2\hat { i } +5\hat { j } -3\hat { k } \) and \(\ \vec { n } =6\hat { i } -3\hat { j } +2\hat { k } \) in the formula, we get
\(\delta =\frac { |\vec { u } .\vec { n } -p| }{ |\vec { n } | } =\frac { |(2\hat { i } +5\hat { j } -3\hat { k } ).(6\hat { i } -3\hat { j } +2\hat { k } )-5| }{ |6\hat { i } -3\hat { j } +2\hat { k } | } \) = 2 unit.

Here (x₁, y₁, z₁) = (3, -4, -3) and direction ratios of the given straight line are (a, b, c) = (-4, -7, 12). 
Direction ratios of the normal to the given plane are (A, B, C) = (5, -1, 1).
We observe that, the given point (x₁, y₁, z₁) = (3, 4, -3) satisfies the given plane 5x-y+z = 8
Next, aA+bB+cC = (-4)(5)+(-7)(-1)+(12)(1) = -1 \(\neq \) 0.
So, the normal to the plane is not perpendicular to the line.
Hence, the given line does not lie in the plane.

Vector form of the equation of the plane is
\(\vec { r } .(6\hat { i } +4\hat { j } -3\hat { k } )\) = 12
Let \(\vec { r } =x\hat { i } +y\hat { j } +z\hat { k } \)
\(\Rightarrow (x\hat { i } +y\hat { j } +z\hat { k } ).(6\hat { i } +4\hat { j } -3\hat { k } )=12\)
⇒ 6x + 4y - 3z = 0
Dividing by 12, we get
\(\frac { 6x }{ 12 } +\frac { 4y }{ 12 } +\frac { 3z }{ 12 } =1\)
[\(\because \frac { x }{ a } +\frac { y }{ b } +\frac { z }{ c } =1\) is the equation of the plane in intercept form]
⇒ \(\frac { x }{ 2 } +\frac { y }{ 3 } +\frac { z }{ -4 } =1\)
∴ The x-intercepts of the plane is 2, y intercept is 3 and z-intercept is -4.

If \(\vec { r } =x\hat { i } +y\hat { j } +z\hat { k } \) is the position vector of an arbitrary point (x, y, z) on the plane, then the given equation can be written as \((x\hat { i } +y\hat { j } +z\hat { k } ).(3\hat { i } -4\hat { j } +3\hat { k } )=-8\) or \((x\hat { i } +y\hat { j } +z\hat { k } ).(-3\hat { i } +4\hat { j } -3\hat { k } )=8\). 
That is, \(\hat { r } .(-3\hat { i } +4\hat { j } -3\hat { k } )=8\) which is the vector equation of the given plane in standard form.

Given lines are \(\vec { r } =(4\hat { i } -\hat { j } )+t(\hat { i } +2\hat { j } -2\hat { k } )\) \([\vec { r } =\vec { a } +t\vec { b } ]\)
∴ \(\vec { b } =\hat { i } +2\hat { j } -2\hat { k } \)
and \(\vec { r } =(\hat { i } -2\hat { j } +4\hat { k } )+s(-\hat { i } -2\hat { j } +2\hat { k } )\)
∴ \(\vec { d } =-\hat { i } -2\hat { j } +2\hat { k } \)
Let θ be the angle between the given lines
Then cos θ = \(\frac { \vec { b } .\vec { d } }{ |\vec { b } ||\vec { d } | } \)
= \(\frac { (\hat { i } +2\hat { j } -2\hat { k } ).(-\hat { i } -2\hat { j } +2\hat { k } ) }{ \sqrt { { 1 }^{ 2 }+{ 2 }^{ 2 }+(-2)^{ 2 } } .\sqrt { (-1)^{ 2 }+{ (-2) }^{ 2 }+{ (2) }^{ 2 } } } \)
= \(\frac { -1-4-4 }{ \sqrt { 9 } .\sqrt { 9 } } =\frac { -9 }{ 9 } \) = -1
∴ cos θ = -1
⇒ cos-1(-1)
⇒ θ = 0

We observe that the straight line \(\frac { x-1 }{ 4 } =\frac { 2-y }{ 6 } =\frac { z-4 }{ 12 } \) is parallel to the vector \(4\hat { i } -6\hat { j } +12\hat { k } \) and the straight line \(\frac { x-3 }{ -2 } =\frac { y-3 }{ 3 } =\frac { 5-z }{ 6 } \) is parallel to the vector \(2\hat { i } +3\hat { j } -6\hat { k } \)
Since \(4\hat { i } -6\hat { j } +12\hat { k } =-2(-2\hat { i } +3\hat { j } -6\hat { k } )\), two vectors are parallel, and hence the two straight lines are parallel.

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

Given \(\vec { a } =\hat { i } -2\hat { j } +3\hat { k } ,\vec { b } =2\hat { i } +\hat { j } +\hat { k } \) and \(\vec { c } =3\hat { i } +2\hat { j } +\hat { k } \)
(i) \((\vec { a } \times \vec { b } )\) = \(\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 1 & -2 & 3 \\ 2 & 1 & -2 \end{matrix} \right| \)
= \(\hat { i } \left| \begin{matrix} -2 & 3 \\ 1 & -2 \end{matrix} \right| -j\left| \begin{matrix} 1 & 3 \\ 2 & -2 \end{matrix} \right| +\hat { k } \left| \begin{matrix} 1 & -2 \\ 2 & 1 \end{matrix} \right| \)
= \(\\ \hat { i } \)(4-3)-\(\\ \hat { j } \)(-2-6)+\(\\ \hat { k } \)(1+4)
= \(\hat { i } +8\hat { j } +5\hat { k } \)
\((\vec { a } \times \vec { b } )\times \vec { c } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 1 & 8 & 5 \\ 3 & 2 & 1 \end{matrix} \right| \)
=\(\hat { i } \left| \begin{matrix} 8 & 5 \\ 2 & 1 \end{matrix} \right| \hat { -j } \left| \begin{matrix} 1 & 5 \\ 3 & 1 \end{matrix} \right| +\hat { k } \left| \begin{matrix} 1 & 8 \\ 3 & 2 \end{matrix} \right| \)
 = \(\\ \hat { i } \)(8-10)-\(\\ \hat { j } \)(1-15)+\(\\ \hat { k } \)(2-24)
= \(-2\hat { i } +14\hat { j } -22\hat { k } \)
(ii) \(\vec { a } \times (\vec { b } \times \vec { c } )\)
\(\vec { b } \times \vec { c } \) = \(\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 2 & 1 & -2 \\ 3 & 2 & 1 \end{matrix} \right| \)
= \(\hat { i } \left| \begin{matrix} 1 & -2 \\ 2 & 1 \end{matrix} \right| \hat { -j } \left| \begin{matrix} 2 & -2 \\ 3 & 1 \end{matrix} \right| +\hat { k } \left| \begin{matrix} 2 & 1 \\ 3 & 2 \end{matrix} \right| \)
= \(\\ \hat { i } \)(1+4)-\(\\ \hat { j } \)(2+6)+\(\\ \hat { k } \)(4-3) = \(5\hat { i } -8\hat { j } +\hat { k } \)
∴ \(\vec { a } \times (\vec { b } \times \vec { c } )=\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 1 & -2 & 3 \\ 5 & -8 & 1 \end{matrix} \right| \)
= \(\hat { i } \left| \begin{matrix} -2 & -3 \\ -8 & 1 \end{matrix} \right| -\hat { j } \left| \begin{matrix} 1 & 3 \\ 5 & 1 \end{matrix} \right| +\hat { k } \left| \begin{matrix} 1 & -2 \\ 5 & -8 \end{matrix} \right| \)
= \(\\ \hat { i } \)(-2-24)-\(\\ \hat { j } \)(1-15)+\(\\ \hat { k } \)(-8-10)
= \(22\hat { i } +14\hat { j } +2\hat { k } \).

  \(\vec { a } \)= \(a\hat { i } +a\hat { j } +c\hat { k } , \vec{b}=\hat { i } +\hat { k } \), \(\vec { c } \)= \(c\hat { i } +c\hat { j } +b\hat { k } \)
Given \(\vec { a } ,\vec { b } \) and \(\vec { c } \) are co-planar
\(\vec { a } .(\vec { b } \times \vec { c } )\) = 0
⇒ \(\left| \begin{matrix} a & a & c \\ 1 & 0 & 1 \\ c & c & b \end{matrix} \right| \) = 0
⇒ \(a\left| \begin{matrix} 0 & 1 \\ c & b \end{matrix} \right| -a\left| \begin{matrix} 1 & 1 \\ c & b \end{matrix} \right| +c\left| \begin{matrix} 1 & 0 \\ c & c \end{matrix} \right| \) = 0
⇒ a(0-c)-a(b-c)+c(c-0) = 0
\(\Rightarrow-\not a c-a b+\not a c+c^{2}=0\)
⇒ c² = ab ⇒ c =\(\sqrt { ab } \).
Hence c is the geometric mean of a and b.

Given \(\vec { a } =\hat { i } -\hat { k } ,\vec { b } =x\hat { i } +\hat { j } +(1-x)\hat { k } ,\vec { c } =y\hat { i } +x\hat { j } +(1+x+y)\hat { k } \)
\([\vec { a } ,\vec { b } \vec { c } ]=\vec { a } .(\vec { b } \times \vec { c } )=\left| \begin{matrix} 1 & 0 & -1 \\ x & 1 & 1-x \\ y & x & 1+x-y \end{matrix} \right| \)
= \(1\left| \begin{matrix} 1 & 1-x \\ x & 1+x-y \end{matrix} \right| +0-y\left| \begin{matrix} x & y \\ y & x \end{matrix} \right| \)
= [(1+x-y)-x(1-x)]-[x²-y]
\(=1+\not x-\not y-\not x+\not x^{2}-\not x^{x}+\not y\)
= 1
∴ \([\vec { a } \vec { b } \vec { c } ]\) = 1 for all values of x and y
∴ \([\vec { a } \vec { b } \vec { c } ]\) depends on neither x nor y.

Given \(\vec { a } =\hat { i } +\hat { j } +\hat { k } \), \(\vec { b } =\hat { i } \), \(\vec { c } ={ c }_{ 1 }\hat { i } +{ c }_{ 2 }\hat { j } +{ c }_{ 3 }\hat { k } \)
∴ \(\vec { c } =\hat { i } +2\hat { j } +{ c }_{ 3 }\hat { k } \)
Also, it given that \(\vec { a } ,\vec { b } \) and \(\vec { c } \) are co-planar.
∴ \(\vec { a } .(\vec { b } \times \vec { c } )\)
⇒ \(\left| \begin{matrix} 1 & 1 & 1 \\ 1 & 0 & 0 \\ 1 & 2 & { c }_{ 3 } \end{matrix} \right| \)= 0
⇒ \(1\left| \begin{matrix} 0 & 0 \\ 2 & { c }_{ 3 } \end{matrix} \right| -1\left| \begin{matrix} 1 & 0 \\ 1 & { c }_{ 3 } \end{matrix} \right| +1\left| \begin{matrix} 1 & 0 \\ 1 & 2 \end{matrix} \right| \) = 0
⇒ 1(0-0)-1(c₃-0)+1(2-0) = 0
⇒ 0 - c₃+2 = 0 ⇒ c₃ = 2
∴ c₃ = 2

Let \(\vec { a } \) = \(2\hat { i } +3\hat { j } +\hat { k } \), \(\vec { b } \)= \(\hat { i } -2\hat { j } +2\hat { k } \) and \(\vec { c } \) = \(\hat { 3i } +\hat { j } +3\hat { k } \)
\(\vec { a } ,\vec { b } \) and \(\vec { c } \) are coplanar if  \(\vec { a } .(\vec { b } \times \vec { c } )\)
Consider \(\vec { a } .(\vec { b } \times \vec { c } )\)
= \(\left| \begin{matrix} 2 & 3 & 1 \\ 1 & -2 & 2 \\ 3 & 1 & 3 \end{matrix} \right| =2\left| \begin{matrix} -2 & 2 \\ 1 & 3 \end{matrix} \right| -3\left| \begin{matrix} 1 & 2 \\ 3 & 3 \end{matrix} \right| +1\left| \begin{matrix} 1 & -2 \\ 3 & 1 \end{matrix} \right| \)
= 2 (-6- 2) -3 (3 - 6) + 1(1 + 6)
= 2(-8) - 3(-3) + 1(7)
= -16 + 9 + 7
= -16+16
= 0.
Hence, the given vectors are co-planar.

Let \(\vec { a } =7\hat { i } +\lambda \hat { j } -3\hat { k } ,\vec { b } =\hat { i } +2\hat { j } -\hat { k } \) and \(\vec { c } =-3\hat { i } +7\hat { j } -5\hat { k } \)
∴ volume of the parallelepiped
= \(\vec { a } .(\vec { b } \times \vec { c } )\)
Given \(\vec { a } .(\vec { b } \times \vec { c } )\) = 90
⇒ \(\left| \begin{matrix} 7 & \lambda & -3 \\ 1 & 2 & -1 \\ -3 & 7 & 5 \end{matrix} \right| \) = 90
⇒ \(-6\left| \begin{matrix} 2 & -1 \\ 7 & 5 \end{matrix} \right| -\lambda \left| \begin{matrix} 1 & -1 \\ -3 & 5 \end{matrix} \right| -3\left| \begin{matrix} 1 & 2 \\ -3 & 7 \end{matrix} \right| \) = 90
⇒ 7(10+7)-λ(5-3)-3(7+6) = 90
⇒ 7(17)-λ(2)-3(13) = 90
⇒ 119-2λ-39 = 90
⇒ 119-39-90 = 2λ
⇒ -10 = 2λ
⇒ λ = -5

Let \(\vec { a } =-6\hat { i } +14\hat { j } +10\hat { k } \), \(\vec { b } =14\hat { i } -10\hat { j } -6\hat { k } \) and \(\vec { c } =2\hat { i } +4\hat { j } -2\hat { k } \)
Volume of the parallelepiped having \(\vec { a } ,\vec { b } \) and \(\vec { c } \) as its co-terminus edges is \(\vec { a } .(\vec { b } \times \vec { c } )\).
∴ \(\vec { a } .(\vec { b } \times \vec { c } )=\left| \begin{matrix} -6 & 14 & 10 \\ 14 & -10 & -6 \\ 2 & 4 & -2 \end{matrix} \right| \)
= \(-6\left| \begin{matrix} -10 & -6 \\ 4 & -2 \end{matrix} \right| -14\left| \begin{matrix} 14 & -6 \\ 2 & -2 \end{matrix} \right| +10\left| \begin{matrix} 14 & -10 \\ 2 & 4 \end{matrix} \right| \)
= -6(20 + 24) - 14(-28 + 12) + 10(56 + 20)
= -6(44) -14(-16) + 10(76)
= -264 + 224 + 760 = 720.
∴ Volume of the required parallelepiped = 720 cubic units.

Given \(\vec { a } =\hat { i } -2\hat { j } +3\hat { k }, \vec { b } =2\hat { i } +\hat { j } -2\hat { k }, \vec { c } =3\hat { i } +2\hat { j } +\hat { k } \)
∴ \(\vec { a } .(\vec { b } \times \vec { c } )=\left| \begin{matrix} 1 & -2 & 3 \\ 2 & 1 & -2 \\ 3 & 2 & 1 \end{matrix} \right| \)
= \(1\left| \begin{matrix} 1 & -2 \\ 2 & 1 \end{matrix} \right| +2\left| \begin{matrix} 2 & -2 \\ 3 & 1 \end{matrix} \right| +3\left| \begin{matrix} 2 & 1 \\ 3 & 2 \end{matrix} \right| \)
= 1(1+4)+2(2+6)+3(4-3)
= 1(5)+2(8)+3(1)
= 5+16+3 = 24
\(\vec { a } .(\vec { b } \times \vec { c } )\) = 24

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

Here, \(\vec { a } =\hat { i } +\hat { 2j } -\hat { 3k } \), \(\vec { b } =\hat { 2i } -\hat { j } +\hat { 2k } \), \(\vec { c } =\hat { 3i } +\hat { j } -\hat { k } \)
We know that  \(\vec { a } ,\vec { b } ,\vec { c } \)  are coplanar if and only if \([\vec { a } ,\vec { b } ,\vec { c } ]\) = 0. Now, \([\vec { a } ,\vec { b } ,\vec { c } ]\) = \(\left| \begin{matrix} 1 & 2 & -3 \\ 2 & -1 & 2 \\ 3 & 1 & -1 \end{matrix} \right| =0\)
Therefore, the three given vectors are coplanar.

Here A = 0, B = 0, C = 1, D = 4, E = 3, F = 4
B = 0, A = 0 either A or C is 0.
Hence, the given equation represents a parabola.

A = 11, C = -25, D = -44, E = - 50, and F = -256
Here A ≠ C and A and C are of opposite signs. Hence, the given equation represents a hyperbola.

Here A = 1, B = 0, C = 1, D = 1, E = -1
Here A = C and B = 0 there i no xy term.
Hence, the given equation represent a circle.

Here A = 3, C = 2 and F = -14
A ≠ C and A and C are of the same sign.
Hence, the given equation represents an ellipse.

Here A = 3, B = 0, C = 3, D = -4, E = 3 and F = 10
A = C and B = 0 (No xy term)
Hence, the given equations represents a circle.

Here A = 2, B = 0, C = -1, F = -7
Here A ≠ C and A and C are of opposite signs.
Hence the given equation represents a hyperbola.

We know that the volume of the parallelepiped whose coterminus edges are \(\vec { a } ,\vec { b } ,\vec { c } \) is given by |\([\vec { a } ,\vec { b } ,\vec { c } ]\)|. Here, \(\vec { a } =\hat { 2i } -\hat { 3j } +\hat { 4k } ,\vec { b } =\hat { i } +\hat { 2j } -\hat { k } ,\vec { c } =\hat { 3i } -\hat { j } +\hat { 2k } \)
Since \([\vec { a } ,\vec { b } ,\vec { c } ]\)  = \(\left| \begin{matrix} 2 & -3 & 4 \\ 1 & 2 & -1 \\ 3 & -1 & 2 \end{matrix} \right| =-7\) , the volume of the given parallelepiped is \(\left| -7 \right| =7\) cubic units.

By the defination of scalar triple product of three vectors,
We find, \(\hat { a } .(\hat { b } \times \hat { c } )\) = \(\left| \begin{matrix} -3 & -1 & 5 \\ 1 & -2 & 1 \\ 0 & 4 & -5 \end{matrix} \right| =-3\)

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

Reducing 9x²-16y² = 144 to the standard form,
we have, \(\frac { { x }^{ 2 } }{ 16 }- \frac { { y }^{ 2 } }{ 9 } =1\)
With the transverse axis is along x-axis vertices are (−4, 0) and (4, 0); and c² = a²+b² = 16 + 9 = 25, c = 5  
Hence the foci are (−5, 0) and (5, 0) 

Equation of the circle is x² + (y + 2)² = 0
Compare with(x-h)²+(y-k)² = r²
h = 0, k = -2, r² = 0
Centre (h, k) = (0, -2)
radius is 0.

Given equation of the circle is
x² + y² - 5x + 2y - 5 = 0
(i) At (-2, 1), (1) becomes
(-2)² + 1²- 5(-2) + 2(1) - 5
= 4 + 1 + 10 + 2 - 5
= 17 - 5 = 12 > 0
∴ (-2, 1) lies outside the circle.
(ii) At (0, 0), (1) becomes -5 < 0
∴ (0, 0) lies inside the circle.
(iii) At (-4, -3), (1) becomes
(-4)² + (-3)² - 5(-4) + 2(-3) - 5
= 16 + 9 + 20 - 6 - 5
= 45 - 11 = 34 > 0
∴ (-4, -3) lies outside the circle.

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

Taking (x₁, y₁) as (2, 3), we get
x₁² + y₁² + 2gx₁+ 2fy₁+ c = 2² + 3² − 6 × 2 − 8 × 3 + 12 
= 4 + 9 - 12 - 24 + 12
= -11\(<\)0.
Therefore the point (2, 3) lies inside the circle, by theorem.

Equation of the circle with end points of the diameter as (x₁, y₁₎ and (x₂, y₂) given in theorem is
(x−x₁)(x−x₂)+(y−y₁)(y−y₂) = 0
(x+4)(x−1)+(y+2)(y−1) = 0
x² + y² + 3x + y − 6 = 0 which is the required equation of the circle.

Centre of the circle is (3,-2) which lies on x + y − 1 = 0. So the line x + y − 1 = 0 passes through the centre and therefore the line x + y −1 = 0 is a diameter of the circle for all possible values of c .