L.H.S \( =\vec{a} \times(\vec{b} \times \vec{c})+\vec{b} \times(\vec{c} \times \vec{a})+\vec{c} \times(\vec{a} \times \vec{b}) \)
\( =(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{c}+(\vec{b} \cdot \vec{a}) \vec{c}-(\vec{b} \cdot \vec{c}) \vec{a}+(\vec{c} \cdot \vec{b}) \vec{a}-(\vec{c} \cdot \vec{a}) \vec{b} \)
\( =\overrightarrow{0}= \) R.H.S

\( (\vec{a} \times \vec{b}) \times(\vec{c} \times \vec{d}) =\vec{x} \times(\vec{c} \times \vec{d}) \text { where } \vec{x}=\vec{a} \times \vec{b} \)
\(=(\vec{x} \cdot \vec{d}) \vec{c}-(\vec{x} \cdot \vec{c}) \vec{d} \)
\(=\{(\vec{a} \times \vec{b}) \cdot \vec{d}\} \vec{c}-\{(\vec{a} \times \vec{b}) \cdot \vec{c}\} \vec{d} \)
\(=[\vec{a}, \vec{b}, \vec{d}] \vec{c}-[\vec{a}, \vec{b}, \vec{c}] \vec{d} \)
Similarly we can prove other result by taking \(\vec{x}=\vec{c} \times \vec{d}\)

\(\vec{x} \cdot \vec{a}=0 \text { and } \vec{x} \cdot \vec{b}=0\) implies \(\vec{a} \text { and } \vec{b} \) are perpendicular to \(\vec x\)
\(\therefore \vec{a} \times \vec{b}\)  is parallel to \(\vec x\)
\(\therefore \vec{x}=\lambda(\vec{a} \times \vec{b})\)
Now, \(\vec{x} \cdot \vec{c}=0 \Rightarrow \lambda(\vec{a} \times \vec{b}) \vec{c}=0 \Rightarrow\left[\begin{array}{lll} \vec{a} & \vec{b} & \vec{c} \end{array}\right]=0\)
\(\Rightarrow \vec{a}, \vec{b}, \vec{c}\) are coplanar.

\(\overrightarrow{O A}=\hat{i}+3 \hat{j}+\hat{k}, \quad \overrightarrow{O B}=\hat{i}+\hat{j}-\hat{k}, \quad \overrightarrow{O C}=-\hat{i}+\hat{j}+\hat{k}\)
\( \overrightarrow{O D}=2 \hat{i}+2 \hat{j}+\hat{k} \)
\( \overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A}=-2 \hat{j}+2 \hat{k} \)
\( \overrightarrow{A C}=\overrightarrow{O C}-\overrightarrow{O A}=2 \hat{i}-2 \hat{j} \)
\( \overrightarrow{A D}=\overrightarrow{O D}-\overrightarrow{O A}=\vec{i}-\vec{j}-2 \vec{k} \)
\( {[\overrightarrow{A B}, \overrightarrow{A C}, \overrightarrow{A D}]=\left|\begin{array}{lll} 0 & -2 & -2 \\ -2 & -2 & 0 \\ 1 & -1 & -2 \end{array}\right|=0} \)

Let \(\theta\) be the angle between \(\vec{a}\ and\ \vec{b}\)
\( \therefore \vec{a} \times \vec{b} =|\vec{a}||\vec{b}| \sin \theta \hat{n} \)
\(|\vec{a} \times \vec{b}|^{2} =|\vec{a}||\vec{b}| \sin \theta \)
\(|\vec{a} \times \vec{b}|^{2} =|\vec{a}|^{2}|\vec{b}|^{2} \sin ^{2} \theta \)
\((\vec{a} \cdot \vec{b})^{2} =|\vec{a}|^{2}|\vec{b}|^{2} \cos ^{2} \theta \)
\(|\vec{a} \times \vec{b}|^{2}+(\vec{a} \cdot \vec{b})^{2} =|\vec{a}|^{2}|\vec{b}|^{2}\left(\sin ^{2} \theta+\cos ^{2} \theta\right) \)
\( =|\vec{a}|^{2}|\vec{b}|^{2} \)

\( |\vec{a} \times \vec{b}|^{2}+(\vec{a} \cdot \vec{b})^{2} =|\vec{a}|^{2}|\vec{b}|^{2} \)
\(|\vec{a} \times \vec{b}|^{2} =|\vec{a}|^{2}|\vec{b}|^{2}-(\vec{a} \cdot \vec{b})^{2} \)
\( =(13)^{2}(5)^{2}-(60)^{2}=625 \)
\(\Rightarrow|\vec{a} \times \vec{b}| =25\)

Let  \(\vec{a}=2 \hat{i}-\hat{j}+\hat{k} ; \vec{b}=\hat{i}-3 \hat{j}-5 \hat{k} ; \vec{c}=-3 \hat{i}+4 \hat{j}+4 \hat{k}\)
We see that \(\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}\)
\(\therefore \vec{a}, \vec{b}, \vec{c}\) forms a triangle
Further \(\vec{a} \cdot \vec{b}=(2 \hat{i}-\hat{j}+\hat{k}) \cdot(\hat{i}-3 \hat{j}-5 \hat{k})=2+3-5=0\)
\(\therefore \vec{a} \perp \vec{b}\)
\(\therefore\) The vectors form the sides of a right angled triangle.

\( |\vec{a}+\vec{b}|^{2}=(\vec{a}+\vec{b})^{2}=|\vec{a}|^{2}+|\vec{b}|^{2}+2 \vec{a} \cdot \vec{b} \)............(1)
\( |\vec{a}-\vec{b}|^{2}=(\vec{a}-\vec{b})^{2}=|\vec{a}|^{2}+|\vec{b}|^{2}-2 \vec{a} \cdot \vec{b} \)...........(2)
Adding (1) and (2)
\( |\vec{a}+\vec{b}|^{2}+|\vec{a}-\vec{b}|^{2} =|\vec{a}|^{2}+|\vec{b}|^{2}+2 \vec{a} \cdot \vec{b}+|\vec{a}|^{2}+|\vec{b}|^{2}-2 \vec{a} \cdot \vec{b} \)
\( =2|\vec{a}|^{2}+2|\vec{b}|^{2}=2\left(|\vec{a}|^{2}+|\vec{b}|^{2}\right) \)

\( \cos \theta =\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}=\frac{(\hat{i}-\hat{j}) \cdot(\hat{j}-\hat{k})}{|\hat{i}-\hat{j}||\hat{j}-\hat{k}|} \)
\(\Rightarrow \cos \theta =\frac{(1)(0)+(-1)(1)+(0)(-1)}{\sqrt{2} \times \sqrt{2}} \)
\(\Rightarrow \cos \theta =-\frac{1}{2} \Rightarrow \theta=\frac{2 \pi}{3}\)

Let \(\vec{a}=3 \hat{i}-2 \hat{j}-6 \hat{k} ; \quad \vec{b}=4 \hat{i}-\hat{j}+8 \hat{k}\)
Let '\(\theta\)' be the angle between the vectors
\( \vec{a} \cdot \vec{b} =12+2-48=-34 \)
\(|\vec{a}| =7,|\vec{b}|=9 \)
\(\cos \theta =\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}=\frac{-34}{7 \times 9} \)
\(\theta =\cos ^{-1}\left(-\frac{34}{63}\right)\)

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

Given \(\overset { \rightarrow }{ OA } =6\overset { \wedge }{ i } -7\overset { \wedge }{ j } ,\overset { \rightarrow }{ OB } =16\overset { \wedge }{ i } -29\overset { \wedge }{ j } -4\overset { \wedge }{ k } ,\overset { \rightarrow }{ OC } =3\overset { \wedge }{ i } -6\overset { \wedge }{ j } \) and \(\overset { \rightarrow }{ OD } =2\overset { \wedge }{ i } +5\overset { \wedge }{ j } +10\overset { \wedge }{ k } \)
\(\overset { \rightarrow }{ AB } =\overset { \rightarrow }{ OB } -\overset { \rightarrow }{ OA } =10\overset { \wedge }{ i } -22\overset { \wedge }{ j } -4\overset { \wedge }{ k } \)
\(\overset { \rightarrow }{ AC } =\overset { \rightarrow }{ OC } -\overset { \rightarrow }{ OA } =-6\overset { \wedge }{ i } +10\overset { \wedge }{ j } -6\overset { \wedge }{ k } \)
\(\overset { \rightarrow }{ AD } =\overset { \rightarrow }{ OD } -\overset { \rightarrow }{ OA } =-4\overset { \wedge }{ i } +12\overset { \wedge }{ j } +10\overset { \wedge }{ k } \)
\(\therefore \left[ \overset { \rightarrow }{ AB } \overset { \rightarrow }{ AC } \overset { \rightarrow }{ AD } \right] =\left| \begin{matrix} 10 \\ -6 \\ -4 \end{matrix}\begin{matrix} -22 \\ -10 \\ 12 \end{matrix}\begin{matrix} -4 \\ -6 \\ 10 \end{matrix} \right| \)
= 10 (100 + 75) + 22(-60 -24) -4(-72 +40)
= 1720 - 1848 + 128 = 0
Hence , the given points are coplanar.

Given \(\overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } =0\)
\(\overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } \)= -\(\overset { \rightarrow }{ c } \) ... (1)
Taking cross product with \(\overset { \rightarrow }{ a } \) both sides, we get 
\(\overset { \rightarrow }{ a } \left( \overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } \right) =-\left( \overset { \rightarrow }{ a } \times \overset { \rightarrow }{ c } \right) \)
\(\overset { \rightarrow }{ a } \times \overset { \rightarrow }{ a } +\overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } =\overset { \rightarrow }{ c } \times \overset { \rightarrow }{ a } \)
\(\left[ \because -\left( \overset { \rightarrow }{ a } \times \overset { \rightarrow }{ c } \right) =\overset { \rightarrow }{ c } \times \overset { \rightarrow }{ a } \right] \)
\(\Rightarrow \overset { \rightarrow }{ a } \times \overset { \rightarrow }{ b } =\overset { \rightarrow }{ c } \times \overset { \rightarrow }{ a } \left( \because \overset { \rightarrow }{ a } \times \overset { \rightarrow }{ a } =\overset { \rightarrow }{ 0 } \right) \)
Taking cross product with \(\overset { \rightarrow }{ b } \) both sides, we get
\(\overset { \rightarrow }{ b } \times \left( \overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } \right) =\overset { \rightarrow }{ b } \times \left( -\overset { \rightarrow }{ c } \right) \)
\(\overset { \rightarrow }{ b } \times \overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } \times \overset { \rightarrow }{ b } =-\overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } \)
\(-\overset { \rightarrow }{ a } \times \overset { \rightarrow }{ b } =-\overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } \)
\(\overset { \rightarrow }{ a } \times \overset { \rightarrow }{ b } =\overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } \)
From (2) and (3) we get
\(\overset { \rightarrow }{ a } \times \overset { \rightarrow }{ b } =\overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } =\overset { \rightarrow }{ c } \times \overset { \rightarrow }{ a } \)

Let AOB be right angled triangle, right angled at O.
Take O as origin.
Then \(\overset { \rightarrow }{ OA } =\overset { \rightarrow }{ a } ,\overset { \rightarrow }{ OB } =\overset { \rightarrow }{ b } \)
\(\therefore \overset { \rightarrow }{ AB } =\overset { \rightarrow }{ OB } -\overset { \rightarrow }{ OA } =\overset { \rightarrow }{ b } -\overset { \rightarrow }{ a } \)
\(\therefore \overset { \rightarrow }{ OA } \bot \overset { \rightarrow }{ OA } =\overset { \rightarrow }{ a } \bot \overset { \rightarrow }{ b } \Rightarrow \overset { \rightarrow }{ a } .\overset { \rightarrow }{ b } =0\)
\(\therefore { AB }^{ 2 }={ \left| \overset { \rightarrow }{ AB } \right| }^{ 2 }=\overset { \rightarrow }{ AB } .\overset { \rightarrow }{ AB } \)
\(=0=\left( \overset { \rightarrow }{ b } -\overset { \rightarrow }{ a } \right) .\left( \overset { \rightarrow }{ b } -\overset { \rightarrow }{ a } \right) \)
\(=\overset { \rightarrow }{ b } .\overset { \rightarrow }{ b } -\overset { \rightarrow }{ b } .\overset { \rightarrow }{ a } -\overset { \rightarrow }{ a } .\overset { \rightarrow }{ b } +\overset { \rightarrow }{ a } .\overset { \rightarrow }{ a } \)
\(={ \left| \overset { \rightarrow }{ b } \right| }^{ 2 }-2\overset { \rightarrow }{ a } .\overset { \rightarrow }{ b } +{ \left| \overset { \rightarrow }{ a } \right| }^{ 2 }\)
= OB² - 0 + OA²    \(\left[ \because \overset { \rightarrow }{ a } .\overset { \rightarrow }{ b } =0 \right] \)
⇒ AB² = OA² + OB²
Hence the Pythagoras theorem

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