Consider \(\overrightarrow {a}+ \overrightarrow {b}+\overrightarrow {c}=\)\(\overrightarrow{BC}+\overrightarrow{CA}+\overrightarrow{AB}=\overrightarrow{BA}+\overrightarrow{AB}=\overrightarrow{0}\)
\(\overrightarrow {a}+ \overrightarrow {b}=-\overrightarrow {c}\)
\(\overrightarrow {a}\times \overrightarrow {b}=\overrightarrow {a}\times (\overrightarrow {a}+\overrightarrow {b})\)        \([\because \overrightarrow {a}\times (\overrightarrow {a}+\overrightarrow {b})=\overrightarrow {a}\times \overrightarrow {a}+\overrightarrow {a}\times \overrightarrow {b}=0+\overrightarrow {a}\times \overrightarrow {b}\)\(=0+\overrightarrow {a}\times \overrightarrow {b}=\overrightarrow {a}\times \overrightarrow {b}]\)
 \(=\overrightarrow {a}\times (-\overrightarrow {c})+-\overrightarrow {a}\times \overrightarrow {c}\)
\(\overrightarrow {a}\times\overrightarrow {b} =\overrightarrow {c}\times \overrightarrow {a}\) ....(1)
and \(\overrightarrow {b}\times\overrightarrow {c} =\overrightarrow {b}\times (\overrightarrow {b}+\overrightarrow {c})=\overrightarrow {b}\times (\overrightarrow {a})\)
and \(\overrightarrow {b}\times\overrightarrow {c} =-\overrightarrow {b}\times\overrightarrow {a} =\overrightarrow {a}\times \overrightarrow {b}\).....(2)
From (1)and (2), \(\overrightarrow {a}\times\overrightarrow {b} =\overrightarrow {b}\times\overrightarrow {c} =\overrightarrow {c}\times \overrightarrow {a}\)

Given \(\overrightarrow { a } \) is perpendicular to both \(\overrightarrow { b } \)  and \(\overrightarrow { c } \)
\(\overrightarrow { a } =\lambda (\overrightarrow { b } \times \overrightarrow { c } )\) for some scalar \(\lambda \)
\(\Rightarrow \quad 1={ |\overrightarrow { a } | }^{ 2 }={ \lambda }^{ 2 }|\overrightarrow { b } \times \overrightarrow { c } |^{ 2 }\quad \Rightarrow 1={ \lambda }^{ 2 }[|{ \overrightarrow { b } | }^{ 2 }{ |\overrightarrow { c } | }^{ 2 }-(\overrightarrow { b } .\overrightarrow { c } )^{ 2 }]\)
\(\Rightarrow 1={ \lambda }^{ 2 }[1(1)-{ |\overrightarrow { b } | }^{ 2 }{ |\overrightarrow { c } | }^{ 2 }-{ cos }^{ 2 }\left( \frac { \pi }{ 6 } \right) ]\quad \Rightarrow 1={ \lambda }^{ 2 }[1-{ cos }^{ 2 }\left( \frac { \pi }{ 6 } \right) ]\)
\(\Rightarrow 1={ \lambda }^{ 2 }[1-{ \left( \frac { \sqrt { 3 } }{ 2 } \right) }^{ 2 }]\ \Rightarrow 1={ \lambda }^{ 2 }[1-\frac { 3 }{ 4 } ]\)
\(\Rightarrow 1={ \lambda }^{ 2 }\left( \frac { 1 }{ 4 } \right) \)
\({ \lambda }^{ 2 }=4\)
\(\lambda =\pm 2\)
\(\therefore \overrightarrow { a } =\pm 2(\overrightarrow { b } \times \overrightarrow { c } )\)

Given \(\left| \vec { a } +\vec { b } \right| =1\)
\({ \left| \vec { a } +\vec { b } \right| }^{ 2 }=1\)
\({ \left| \vec { a } \right| }^{ 2 }+{ \left| \vec { b } \right| }^{ 2 }+2(\vec { a } .\vec { b } )=1\)
1+1+2\({ \left| \vec { a } \right| }{ \left| \vec { b } \right| }\) cos\(\theta\) =1 where \(\theta\) is the angle between \(\vec{a}\)and \(\vec{b}\)
2+2(1)(1)cos\(\theta\) =1
cos\(\theta\) =1-2=-1
cos \(\theta\)=-\(\frac{1}{2}\)
Consider \({ \left| \vec { a } -\vec { b } \right| }^{ 2 }={ \left| \vec { a } \right| }^{ 2 }+{ \left| \vec { b } \right| }^{ 2 }-2(\vec { a } .\vec { b } )=1+1-2\left| \vec { a } \right| \left| \vec { b } \right| cos\theta \)
=2-2(1)(1) \(\left( -\frac { 1 }{ 2 } \right) \)=2+1=3
\(\therefore \left| \vec { a } -\vec { b } \right| =\sqrt { 3 } \)

Given \((\vec { a } +\vec { b } ).\vec { c } =0\Rightarrow \vec { a } .\vec { c } +\vec { b } .\vec { c } =0\)
\((\vec { b } +\vec { c } ).\vec { a } =0\Rightarrow \vec { b } .\vec { a } +\vec { c } .\vec { a } =0\)
\((\vec { c } +\vec { a } ).\vec { b } =0\Rightarrow \vec { c } .\vec { b } +\vec { a } .\vec { b } =0\)
Adding 2\((\vec { a } .\vec { b } +\vec { b } .\vec { c } +\vec { c } .\vec { a } )=0\)
\(\vec { a } .\vec { b } +\vec { b } .\vec { c } +\vec { c } .\vec { a } =0\)  --------(1)
\({ \left| \vec { a } +\vec { b } +\vec { c } \right| }^{ 2 }={ \left| \vec { a } \right| }^{ 2 }+{ \left| \vec { b } \right| }^{ 2 }+{ \left| \vec { c } \right| }^{ 2 }+2(\vec { a } .\vec { b } +\vec { b } .\vec { c } +\vec { c } .\vec { a } )\)
=9+16+24+2(0)=49
\(\Rightarrow { \left| \vec { a } +\vec { b } +\vec { c } \right| }=7\)

Given \(\vec{a}=2\hat{i}+\hat{j}-2\hat{k}\)
\(\left| \vec { a } \right| =\sqrt { { 2 }^{ 2 }+{ 1 }^{ 2 }+({ -2) }^{ 2 } } =\sqrt { 4+1+4 } =\sqrt { 9 } \)=3
\(\vec { b } =\hat { i } +\hat { j } \)
\(\left| \vec { b } \right| =\sqrt { { 1 }^{ 2 }+{ 1 }^{ 2 } } =\sqrt { 2 } \)
\(\left| \vec { c } -\vec { a } \right| =2\sqrt { 2 } \)
\({ \left| \vec { c } -\vec { a } \right| }^{ 2 }={ (2\sqrt { 2 } ) }^{ 2 }\)
\(\therefore { \left| \vec { c } \right| }^{ 2 }+{ \left| \vec { a } \right| }^{ 2 }-2(\vec { c } .\vec { a } )=8\)
\({ \left| \vec { c } \right| }^{ 2 }+9-2\left| \vec { c } \right| =8\quad [\therefore \left| \vec { a } \right| =3,\vec { c } .\vec { a } =\left| \vec { c } \right| ]\)
\({ \left| \vec { c } \right| }^{ 2 }+2\left| \vec { c } \right| +1=0\)
\({ [\left| \vec { c } \right| -1] }^{ 2 }=0\)
\({ \left| \vec { c } \right| -1 }=0\)  \(\Rightarrow { \left| \vec { c } \right| }=0\)
Also \(\vec { a } \times \vec { b } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 2 & 1 & -2 \\ 1 & 1 & 0 \end{matrix} \right| \begin{matrix} =\hat { i } (0+2)-\hat { j } (0+2)+\hat { k } (2-1) \\ =2\hat { i } -2\hat { j } +\hat { k } \end{matrix}\)
\(\left| \vec { a } \times \vec { b } \right| =\sqrt { { 2 }^{ 2 }+{ (-2) }^{ 2 }+{ 1 }^{ 2 } } =\sqrt { 4+4+1 } =\sqrt { 9 } =3\)
\(\therefore \left| (\vec { a } \times \vec { b } ).\vec { c } \right| =\left| \vec { a } \times \vec { b } \right| \left| \vec { c } \right| sin30\) [angle between \(\vec{a} \times \vec{b}\) and \(\vec{c}\) is 30^o]
=3(1)\(\left( \frac { 1 }{ 2 } \right) =\left( \frac { 3 }{ 2 } \right) \)

\(a^2=|\overrightarrow{a}|^2=|\overrightarrow{BC}|^2=\overrightarrow{BC}.\overrightarrow{BC}\)
\(=-(\overrightarrow{CA}+\overrightarrow{AB}).\overrightarrow{BC}\)    \([\because \overrightarrow{BC}=\overrightarrow{BA}+\overrightarrow{CA}=\overrightarrow{BA}-\overrightarrow{CA}=-\overrightarrow{AB}-\overrightarrow{CA}]\)
\(=-\overrightarrow{CA}+.\overrightarrow{BC}-\overrightarrow{AB}.\overrightarrow{BC}=-|\overrightarrow{CA}||\overrightarrow{BC}|\) cos of angle between \(\overrightarrow{CA}\) and \(\overrightarrow{BC}\)\(-|\overrightarrow{AB}||\overrightarrow{BC}|\)cos of angle between \(\overrightarrow{AB}\) and\(\overrightarrow{BC}\)
= - ba cos (180 - C) - ca cos (180 - B)
a²= ab cos C + ac cos B [\(\therefore\) cos (180 - C) = - cos C;
a²= a (b cos C + c cos B)        cos (180 - B) = - cos B]
a = b cos C + c cos B
Hence proved.

Since \(\overrightarrow { c } \) is co-planar with \(\overrightarrow { a } \) and \(\overrightarrow { b } \)
Let \(\overrightarrow { c } =x\overrightarrow { a } +y\overrightarrow { b } \)  where x, y are sclars.
\(\Rightarrow \overrightarrow { c } =x(\hat { i } +\hat { j } +2\hat { k } )+y(\hat { i } +2\hat { j } +\hat { k } )\)
\(\Rightarrow \overrightarrow { c } =\hat { i } (x+y)+\hat { j } (x+2y)+\hat { k } (2x+y)\)
\(\therefore |\overrightarrow { c } |=1\Rightarrow \sqrt { { (x+y) }^{ 2 }+{ (x+2y) }^{ 2 }+{ (2x+y) }^{ 2 } } =1\) ....(1)
Also, \(\overrightarrow { c } \) is perpendicular to \(\hat { i } +\hat { j } +\hat { k } \)
\(\Rightarrow \overrightarrow { c } .(\hat { i } +\hat { j } +\hat { k } )=0\)
\(\\ \\ \\ \Rightarrow \) (x+y)(1)+(x+2y)1+(2x+y)1 = 0
\(\Rightarrow \) 4x+4y = 0 \(\Rightarrow \) x = -y
Substituting y = -x in (1) we get,
\(\sqrt { 0+{ (-x) }^{ 2 }+{ (x) }^{ 2 } } =1\Rightarrow \sqrt { { 2x }^{ 2 } } =1\)
\(\Rightarrow{ 2x }^{ 2 }=1\Rightarrow { x }^{ 2 }=\frac { 1 }{ 2 } \Rightarrow x=\pm \frac { 1 }{ \sqrt { 2 } } \quad \)
\(\therefore\) when \(x=\frac { 1 }{ \sqrt { 2 } } ,y=-\frac { 1 }{ \sqrt { 2 } } \) and when \(x=-\frac { 1 }{ \sqrt { 2 } } ,y=\frac { 1 }{ \sqrt { 2 } } \)
\(\therefore x=\frac { 1 }{ \sqrt { 2 } } \) and \(y=-\frac { 1 }{ \sqrt { 2 } } \Rightarrow \overrightarrow { c } =-\frac { \hat { j } }{ \sqrt { 2 } } +\frac { \hat { k } }{ \sqrt { 2 } } \)
when \(x=-\frac { 1 }{ \sqrt { 2 } } \)  , and \(y=\frac { 1 }{ \sqrt { 2 } } \Rightarrow \overrightarrow { c } =\frac { \hat { j } }{ \sqrt { 2 } } -\frac { \hat { k } }{ \sqrt { 2 } } \)
\(\therefore\) \(\overrightarrow { a } .\overrightarrow { c } =0+\frac { 1 }{ \sqrt { 2 } } -\frac { 2 }{ \sqrt { 2 } } =\frac { -1 }{ \sqrt { 2 } } <0\)
\(\overrightarrow { c } \) makes obtuse angle with \(\overrightarrow { a } \) means.
\(\overrightarrow { c } =\frac { \hat { j } -\hat { k } }{ \sqrt { 2 } } \)

Let \(\overrightarrow{BC}=\overrightarrow{a},\overrightarrow{AC}=\overrightarrow{b},\overrightarrow{BA}=\overrightarrow{c}\)

Then \(|\overrightarrow{a}|=a,|\overrightarrow{b}|=b,\) and \(|\overrightarrow{c}|=c\)
Since \(\overrightarrow{BC}=\overrightarrow{BA}+\overrightarrow{AC}\)
We have \(\overrightarrow{a}=\overrightarrow{c}+\overrightarrow{b}\) and angle between \(\overrightarrow{c}\) and \(\overrightarrow{b}\) is (180-A)
\(=|\overrightarrow{a}|^2=|\overrightarrow{c}+\overrightarrow{b}|^2=|\overrightarrow{c}|^2+|\overrightarrow{b}|^2+2|\overrightarrow{c}||\overrightarrow{b}|cos (180-A)\)
\(=c^2+b^2+2cb \ cos (180-A)\)
\(=b^2+c^2-2cb \ cos A\)    \([\because cos(180-A)=-cos \ A]\)

Let OABCDEFG be a unit cube.
Keeping O as origin.
Let \(\overrightarrow{OA}=\hat{i},\overrightarrow{OC}=\hat{j}\) and \(\overrightarrow{OG}=\hat{k}\)
Consider the diagonals OE and BG.

\(\overrightarrow{OE}=\overrightarrow{OB}+\overrightarrow{BE}=\overrightarrow{OA}+\overrightarrow{AB}+\overrightarrow{BE}\)
\(=\overrightarrow{OA}+\overrightarrow{OC}+\overrightarrow{OG}=\hat{i}+\hat{j}+\hat{k}\)       \([\because \overrightarrow{AB}= \overrightarrow{OC}, \overrightarrow{BE}= \overrightarrow{OG}]\)
and  \(\overrightarrow{GB}=\overrightarrow{GO}+\overrightarrow{OB}=-\hat{k}+\overrightarrow{OA}+\overrightarrow{AB}=\)\(\hat{i}+\hat{j}-\hat{k}\)
Let \(\theta\) be the smaller angle between the diagonals OE and GB, then
cos \(\theta\)=\({\overrightarrow{OE}.\overrightarrow{GB}\over|\overrightarrow{OE}||\overrightarrow{GB}|}=\)\({1(1)+1(1)+1(-1)\over \sqrt{1^2+1^2+1^2}.\sqrt{1^2+1^2+(-1)^2}}={2-1\over \sqrt{3}\sqrt{3}}={1\over3}\)
Thus \(\theta=\)cos^-1 \(({1\over3})\)

Given \(|\overrightarrow{a}|=2,|\overrightarrow{b}|=3,|\overrightarrow{c}|=4,\)and \(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}\Rightarrow \overrightarrow{a}+\overrightarrow{b}=-\overrightarrow{c}\)
\(\therefore |\overrightarrow{a}+\overrightarrow{b}|^2=|\overrightarrow{c}|^2\)
\(\Rightarrow |\overrightarrow{a}|^2+|\overrightarrow{b}|^2+2(\overrightarrow{a}.\overrightarrow{b})=|\overrightarrow{c}|^2\)
\(\Rightarrow 4+9+2(\overrightarrow{a}.\overrightarrow{b})=16\)
\(\Rightarrow 13+2(\overrightarrow{a}.\overrightarrow{b})=16\)
\(\Rightarrow 2(\overrightarrow{a}.\overrightarrow{b})=16-13=3\)
\(\Rightarrow \overrightarrow{a}.\overrightarrow{b}={3\over2}\)
\(\Rightarrow4( \overrightarrow{a}.\overrightarrow{b})=4\times {3\over2}=6\).....(1)
Also \(\overrightarrow{b}+\overrightarrow{c}=-\overrightarrow{a}\)
\(|\overrightarrow{b}+\overrightarrow{c}|^2=|-\overrightarrow{a}|^2\)
\(|\overrightarrow{b}|^2+|\overrightarrow{c}|^2+2(\overrightarrow{b}.\overrightarrow{c})=|-\overrightarrow{a}|^2\)
\(9+16+2(\overrightarrow{b}.\overrightarrow{c})=4\)
\(25+2(\overrightarrow{b}.\overrightarrow{c})=4\)
\(2(\overrightarrow{b}.\overrightarrow{c})=4-25=-21\)
\((\overrightarrow{b}.\overrightarrow{c})={-21\over 2}\)
\(3(\overrightarrow{b}.\overrightarrow{c})=3({-21\over 2})={-63\over2}\)..(2)
Also, \(\overrightarrow{c}+\overrightarrow{a}=-\overrightarrow{b}\)
\(|\overrightarrow{c}+\overrightarrow{a}|=|-\overrightarrow{b}|\)
\(|\overrightarrow{c}+\overrightarrow{a}|^2=|-\overrightarrow{b}|^2\)
\(|\overrightarrow{c}|^2+|\overrightarrow{a}|^2+2(\overrightarrow{c}.\overrightarrow{a})=|\overrightarrow{b}|^2\)
\(16+4+2(\overrightarrow{c}.\overrightarrow{a})=9\)
\(20+2(\overrightarrow{c}.\overrightarrow{a})=9\)
\(\Rightarrow 2(\overrightarrow{c}.\overrightarrow{a})=9-20=-11\)
\((\overrightarrow{c}.\overrightarrow{a})={-11\over2}\)
\(\therefore 3(\overrightarrow{c}.\overrightarrow{a})=3({-11\over2})={-33\over2}\)....(3)
Adding (1), (2) and (3) we get,
\(4\overrightarrow{a}.\overrightarrow{b}+3\overrightarrow{b}.\overrightarrow{c}+3\overrightarrow{c}.\overrightarrow{a}=\)\(6-{63\over2}-{33\over2}={12-63-33\over 2}={12-96\over2}={-84\over2}=-42\)
\(\therefore 4\overrightarrow{a}.\overrightarrow{b}+3\overrightarrow{b}.\overrightarrow{c}+3\overrightarrow{c}.\overrightarrow{a}=-42\)