The normal vector to the planes
x + 2y + 3z - 7 = 0, 2x - 3y + 4z = 0 are 
\(\overset { \rightarrow }{ b } =\overset { \wedge }{ i } +2\overset { \wedge }{ j } +3\overset { \wedge }{ k } \) and \(\overset { \rightarrow }{ c } =2\overset { \wedge }{ i } -3\overset { \wedge }{ j } +4\overset { \wedge }{ k } \)
∴ The required planes passes through the point \(\overset { \rightarrow }{ a } =\overset { \wedge }{ i } +\overset { \wedge }{ j } +\overset { \wedge }{ k } \) and parallel to two vector 5 namely \(\overset { \rightarrow }{ b } \) and \(\overset { \rightarrow }{ c } \) 
∴ The Parametric form of vectors equation of the plans is \(\overset { \rightarrow }{ r } =\overset { \rightarrow }{ a } +s\overset { \rightarrow }{ b } +t\overset { \rightarrow }{ c } \) s, t ∈ R
\(\overset { \rightarrow }{ r } =\left( \overset { \rightarrow }{ i } +\overset { \rightarrow }{ j } -\overset { \rightarrow }{ k } \right) +s\left( \overset { \rightarrow }{ i } +2\overset { \rightarrow }{ j } +3\overset { \rightarrow }{ k } \right) +t\left( 2\overset { \rightarrow }{ i } -3\overset { \rightarrow }{ j } +4\overset { \rightarrow }{ k } \right) ,\)
Cartesian equation is \(\left| \begin{matrix} x-{ x }_{ 1 } \\ { b }_{ 1 } \\ { c }_{ 1 } \end{matrix}\begin{matrix} y-{ { y }_{ 1 } } \\ { b }_{ 2 } \\ { c }_{ 2 } \end{matrix}\begin{matrix} z-{ { z }_{ 1 } } \\ { b }_{ 3 } \\ { c }_{ 3 } \end{matrix} \right| =0\)
\(\Rightarrow \left| \begin{matrix} x-1 \\ 1 \\ 2 \end{matrix}\begin{matrix} y-1 \\ 2 \\ -3 \end{matrix}\begin{matrix} z+1 \\ 3 \\ 4 \end{matrix} \right| =0\)
⇒ (x - 1) (8 + 9) - (y - 1)(4 - 6) + (z + 1)(-3 -4) = 0
⇒ 17 (x - 1) +2 (y - 1) -7 (z + 1) = 0
⇒ 17x - 17 + 2y - 2 - 7z - 7 = 0
⇒ 17x + 2y - 7z - 26 = 0

Given \(\overset { \rightarrow }{ OA } =2\overset { \wedge }{ i } -\overset { \wedge }{ j } +\overset { \wedge }{ k } ,\overset { \rightarrow }{ OB } =\overset { \wedge }{ i } -3\overset { \wedge }{ j } -5\overset { \wedge }{ k } ,\) and \(\overset { \rightarrow }{ OC } =3\overset { \wedge }{ i } -4\overset { \wedge }{ j } +4\overset { \wedge }{ k } \)

\(\therefore \overset { \rightarrow }{ AB } =\overset { \rightarrow }{ OB } -\overset { \rightarrow }{ OA } =-\overset { \wedge }{ i } -2\overset { \wedge }{ j } -6\overset { \wedge }{ k } \)
\(\overset { \rightarrow }{ BC } =\overset { \rightarrow }{ OC } -\overset { \rightarrow }{ OB } =2\overset { \wedge }{ i } -\overset { \wedge }{ j } +\overset { \wedge }{ k } \)
\(\overset { \rightarrow }{ CA } =\overset { \rightarrow }{ OA } -\overset { \rightarrow }{ OC } =\overset { \wedge }{ -i } +3\overset { \wedge }{ j } +5\overset { \wedge }{ k } \)
\(\overset { \rightarrow }{ AB } +\overset { \rightarrow }{ BC } +\overset { \rightarrow }{ CA } =\overset { \rightarrow }{ 0 } \)
Also, \(\overset { \rightarrow }{ BC } \). \(\overset { \rightarrow }{ CA } \)=\(\left( 2\overset { \wedge }{ i } -\overset { \wedge }{ j } +\overset { \wedge }{ k } \right) .\left( \overset { \wedge }{ -i } +3\overset { \wedge }{ j } +5\overset { \wedge }{ k } \right) \)
= -2 -3 + 5 = 0
\(\Rightarrow \overset { \rightarrow }{ BC } .\overset { \rightarrow }{ CA } \Rightarrow \angle BCA=\frac { \pi }{ 2 } \)
Hence, ABC is a right angled triangle.
\(\cos { A } =\frac { \overset { \rightarrow }{ AB } .\overset { \rightarrow }{ AC } }{ \left| \overset { \rightarrow }{ AB } \right| \left| \overset { \rightarrow }{ AC } \right| } \)
\(=\frac { \left( -\overset { \wedge }{ i } -2\overset { \wedge }{ j } -6\overset { \wedge }{ k } \right) .\left( \overset { \wedge }{ -i } -3\overset { \wedge }{ j } -5\overset { \wedge }{ k } \right) }{ \sqrt { 1+4+36 } .\sqrt { 1+9+25 } } \)
\(=\frac { 35 }{ \sqrt { 41 } .\sqrt { 35 } } =\sqrt { \frac { 35 }{ 41 } } =A={ Cos }^{ -1 }\left( \sqrt { \frac { 35 }{ 41 } } \right) \)
\(\cos { B } =\frac { \overset { \rightarrow }{ BA. } \overset { \rightarrow }{ BC } }{ \left| \overset { \rightarrow }{ BA } \right| \left| \overset { \rightarrow }{ BC } \right| } =\frac { \left( \overset { \wedge }{ i } +2\overset { \wedge }{ j } +6\overset { \wedge }{ k } \right) .\left( 2\overset { \wedge }{ i } -\overset { \wedge }{ j } +\overset { \wedge }{ k } \right) }{ \sqrt { 1+4+36 } +\sqrt { 4+1+1 } } \)
\(=\sqrt { \frac { 6 }{ 41 } } \Rightarrow B={ Cos }^{ -1 }\left( \sqrt { \frac { 6 }{ 41 } } \right) \)

By definition,
\(\vec { a } \times \vec { b } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 1 & -1 & 0 \\ 1 & -1 & -4 \end{matrix} \right| =4\hat { i } +4\hat { j } ,\vec { c } \times \vec { d } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 0 & 3 & -1 \\ 2 & 5 & 1 \end{matrix} \right| =8\hat { i } -2\hat { j } -6\hat { k } \)
\((\vec { a } \times \vec { b } )(\vec { c } \times \vec { d } )=\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 4 & 4 & 0 \\ 8 & -2 & -6 \end{matrix} \right| =-24\hat { i } +24\hat { j } -40\hat { k } \) ..............(1)
On the other hand, we have
\([\vec { a }, \vec { b }, \vec { d } ]\vec { c } -[\vec { a } ,\vec { b } ,\vec { c } ]\vec { d } =28(3\vec { j } -\vec { k } )-12(2\hat { i } +5\hat { j } +\hat { k } )=-24\hat { i } +24\hat { j } -40\hat { k } \) ..............(2)
Therefore, from equations (1) and (2), identity (i) is verified.
The verification of identity (ii) is left as an exercise to the reader

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

Let A be the origin, \(\vec { b } \) be the position vector of B and \(\vec {c } \) be the position vector of C .
Now D is the midpoint of BC , and so the position vector of D \(\frac{\vec{b}+\vec{c}}{2}\). There, we get

\({ \left| \vec { AD } \right| }^{ 2 }=\vec { AD } .\vec { AD } \)= \(\left( \frac { \vec { b } +\vec { c } }{ 2 } \right) .\left( \frac { \vec { b } +\vec { c } }{ 2 } \right) \)= \(\frac { 1 }{ 4 } ({ \left| \vec { b } \right| }^{ 2 }+{ \left| \vec { c } \right| }^{ 2 }+2\overrightarrow { b } .\vec { c } )\)....(1)
Now, \(\vec { BD } =\vec { AD } -\vec { AB } \) = \(\frac { \vec { b } +\vec { c } }{ 2 } -\vec { b }=\frac { \vec {c } -\vec { b} }{ 2 }\)
Then, we get, =\({ \left| \vec { BD } \right| }^{ 2 }=\vec { BD } .\vec {BD } \) = \(\left( \frac { \vec { c } -\vec { b} }{ 2 } \right) .\left( \frac { \vec { c } -\vec { b} }{ 2 } \right) \) = \(\frac { 1 }{ 4 } ({ \left| \vec { b } \right| }^{ 2 }+{ \left| \vec { c } \right| }^{ 2 }-2\overrightarrow { b } .\vec { c } )\)....(2)
Now, adding (1) and (2), we get
Therefore, \({ \left| \vec { AD } \right| }^{ 2 }+{ \left| \vec { BD } \right| }^{ 2 }=\frac { 1 }{ 4 } ({ \left| \vec { b } \right| }^{ 2 }+{ \left| \vec { c } \right| }^{ 2 }+2\overrightarrow { b } .\vec { c } )+\frac { 1 }{ 4 } ({ \left| \vec { b } \right| }^{ 2 }+{ \left| \vec { c } \right| }^{ 2 }-2\overrightarrow { b } .\vec { c } )=\frac { 1 }{ 2 } ({ \left| \vec { b } \right| }^{ 2 }+{ \left| \vec { c } \right| }^{ 2 })\)
⇒ \({ \left| \vec { AD } \right| }^{ 2 }+{ \left| \vec { BD } \right| }^{ 2 }=\frac { 1 }{ 2 } ({ \left| \vec { AB } \right| }^{ 2 }+{ \left| \vec { AC } \right| }^{ 2 })\)
Hence, \({ \left| \vec { AB } \right| }^{ 2 }+{ \left| \vec { AC } \right| }^{ 2 }=2({ \left| \vec { AD } \right| }^{ 2 }+{ \left| \vec { BD } \right| }^{ 2 })\)

From the line \(\frac { x-1 }{ 3 } =\frac { y+1 }{ 2 } =\frac { z-1 }{ 5 } \)
(x₁, y₁, z₁) is (1, -1, 1)
(l₁, m₁, n₁) is (3, 2, 5)
From the line \(\frac { x+2 }{ 4 } =\frac { y-1 }{ 3 } =\frac { z+1 }{ -2 } \)
we get, (x₂, y₂, z₂) is (-2, 1, -1)
(l₂, m₂, n₂) is 4, 3, -2
The Condition for intersecting lines is 
\(\left| \begin{matrix} { x }_{ 2 }-{ x }_{ 1 } \\ { l }_{ 1 } \\ { l }_{ 2 } \end{matrix}\begin{matrix} { y }_{ 2 }-{ y }_{ 1 } \\ { m }_{ 1 } \\ { m }_{ 2 } \end{matrix}\begin{matrix} { z }_{ 2 }-{ z }_{ 1 } \\ { n }_{ 1 } \\ { n }_{ 2 } \end{matrix} \right| =0\)
\(\Rightarrow \left| \begin{matrix} -2-1 \\ 3 \\ 4 \end{matrix}\begin{matrix} 1+1 \\ 2 \\ 3 \end{matrix}\begin{matrix} -1-1 \\ 5 \\ -2 \end{matrix} \right| \)
\(\Rightarrow \left| \begin{matrix} -3 \\ 3 \\ 4 \end{matrix}\begin{matrix} 2 \\ 2 \\ 3 \end{matrix}\begin{matrix} -2 \\ 5 \\ -2 \end{matrix} \right| \)
= -3 (-4 -15) -2 (-6 -20) -2 (9 - 8)
= -3(-19) - 2(-26) -2 (1)
= 57 + 52 - 2 = 57 + 50
= 107 ≠ 0
Hence the given lines do not intersect

L. H. S = \(\left[ \overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } ,\overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } ,\overset { \rightarrow }{ c } \right] \)
\(=\left( \overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } \right) .\left\{ \left( \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } \right) \times \overset { \rightarrow }{ c } \right\} \)
\(=\left( \overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } \right) .\left\{ \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } \times \overset { \rightarrow }{ c } \right\} \)
\(=\left( \overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } \right) .\left\{ \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } \right\} \ \left[ \because \overset { \rightarrow }{ c } \times \overset { \rightarrow }{ c } =\overset { \rightarrow }{ 0 } \right] \)
\(=\overset { \rightarrow }{ a } .\left( \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } \right) +\overset { \rightarrow }{ b } .\left( \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } \right) +\overset { \rightarrow }{ c } .\left( \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } \right) \)
\(=\left[ \overset { \rightarrow }{ a } \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \right] +0+0\)
\(\left[ \because \left[ \overset { \rightarrow }{ b } \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \right] =\left[ \overset { \rightarrow }{ c } \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \right] =0 \right] \)
\(=\left[ \overset { \rightarrow }{ a } \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \right] \)= R. H. S
Hence proved

The straight line passing through the points (6, 7, 5)A and (8, 10, 6)B is parallel to the vector \(\vec { b } =\vec { AB } =\vec { OB } -\vec { OA } =2\hat { i } +3\hat { j } +\hat { k } \) and the straight line passing through the points C(10, 2, -5) and D(8, 3, -4) is parallel to the vector \(\vec { d } =\vec { CD } =-2\hat { i } +\hat { j } +\hat { k } \). Therefore, the angle between the two straight lines is the angle between the two vectors \(\vec { b } \)and \(\vec { d } \).
Since \(\vec { b } .\vec { d } =(2\hat { i } +3\hat { j } +\hat { k).( } -2\hat { i } +\hat { j } +\hat { k } ) =0\)
the two vectors are perpendicular, and hence the two straight lines are perpendicular. 
Aliter :
We find that direction ratios of the straight line joining the points A(6, 7, 5) and B(8,10, 6) are (b₁, b₂, b₃ ) = (2, 3, 1) and direction ratios of the line joining the points C(10, 2, −5) and D(8, 3, −4) are (d₁, d₂, d₃ ) = (−2, 1, 1). Since  b₁d₁+ b₂d₂ + b₃d₃ = (2)(−2) + (3)(1) + (1)(1) = 0, the two straight lines are perpendicular.

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.

Given \(\overset { \rightarrow }{ a } .\overset { \rightarrow }{ b } =\overset { \rightarrow }{ a } .\overset { \rightarrow }{ c } =0\)] ⇒ \(\overset { \rightarrow }{ a } \) 丄 \(\overset { \rightarrow }{ b } \)and \(\overset { \rightarrow }{ a } \)丄 \(\overset { \rightarrow }{ c } \) 
⇒ \(\overset { \rightarrow }{ a } \) 丄^r to the plane containing \(\overset { \rightarrow }{ b } \)and \(\overset { \rightarrow }{ c } \) 
Also, \(\overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } =\left| \overset { \rightarrow }{ b } \right| \left| \overset { \rightarrow }{ c } \right| \) Since \(\overset { \wedge }{ n } \) [where θ is the angle between \(\overset { \rightarrow }{ b } \)and \(\overset { \rightarrow }{ c } \)]
= 1 \(\times\) 1. sin \(\frac { \pi }{ 6 } \).\(\overset { \rightarrow }{ a } \)
[Since \(\overset { \rightarrow }{ b } \) and \(\overset { \rightarrow }{ c } \) are unit vectors \(\overset { \rightarrow }{ a } \) 丄 both \(\overset { \rightarrow }{ b } \)and \(\overset { \rightarrow }{ c } \)  \(\overset { \rightarrow }{ { n } } \) = \(\overset { \rightarrow }{ a } \)]
\(=\frac { 1 }{ 2 } \overset { \rightarrow }{ a } \)
\( \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } =\frac { 1 }{ 2 } \overset { \rightarrow }{ a } \)
\(\Rightarrow \overset { \rightarrow }{ a } =\pm 2\left( \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } \right) \)

Since the plane passing through the point \(\overset { \rightarrow }{ a } =\overset { \wedge }{ i } -\overset { \wedge }{ j } +2\overset { \wedge }{ k } \) and is normal to the vector \(\overset { \rightarrow }{ n } =2\overset { \wedge }{ i } +3\overset { \wedge }{ j } +2\overset { \wedge }{ k } \)
the vector equation of the plane is \(\overset { \rightarrow }{ r } .\overset { \rightarrow }{ n } =\overset { \rightarrow }{ a } .\overset { \rightarrow }{ n } \)
\(\Rightarrow { \overset { \rightarrow }{ r } }.\left( 2\overset { \wedge }{ i } +3\overset { \wedge }{ j } +2\overset { \wedge }{ k } \right) =\left( \overset { \wedge }{ i } -\overset { \wedge }{ j } +2\overset { \wedge }{ k } \right) .\left( 2\overset { \wedge }{ i } +3\overset { \wedge }{ j } +2\overset { \wedge }{ k } \right) \)
\(\Rightarrow { \overset { \rightarrow }{ r } }.\left( 2\overset { \wedge }{ i } +3\overset { \wedge }{ j } +2\overset { \wedge }{ k } \right) \)= 2 - 3 + 4 = 3
\(\therefore { \overset { \rightarrow }{ r } }.\left( 2\overset { \wedge }{ i } +3\overset { \wedge }{ j } +2\overset { \wedge }{ k } \right) \)= 3

The parametric form of vector equation of a line passing through a point \(\left( \overset { \rightarrow }{ a } \right) \) and parallel to \(\overset { \rightarrow }{ b } \) is
\({ \overset { \rightarrow }{ r } }=\overset { \rightarrow }{ a } +t\overset { \rightarrow }{ b } \), t ∈ R
⇒ Here \(\overset { \rightarrow }{ a } =2\overset { \wedge }{ i } -\overset { \wedge }{ j } +3\overset { \wedge }{ k } \) and \(\overset { \rightarrow }{ b } =2\overset { \wedge }{ i } +\overset { \wedge }{ j } -2\overset { \wedge }{ k } \)
\(\therefore { \overset { \rightarrow }{ r } }=\left( 2\overset { \wedge }{ i } -\overset { \wedge }{ j } +3\overset { \wedge }{ k } \right) +t\left( 2\overset { \wedge }{ i } +\overset { \wedge }{ j } -2\overset { \wedge }{ k } \right) \), t ∈ R which is the required equation of a line.

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.

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