\(a^2=|\overrightarrow{a}|^2=|\overrightarrow{BC}|^2=\overrightarrow{BC}.\overrightarrow{BC}\)
\(=-(\overrightarrow{CA}+\overrightarrow{AB}).\overrightarrow{BC}\)    \([\because \overrightarrow{BC}=\overrightarrow{BA}+\overrightarrow{CA}=\overrightarrow{BA}-\overrightarrow{CA}=-\overrightarrow{AB}-\overrightarrow{CA}]\)
\(=-\overrightarrow{CA}+.\overrightarrow{BC}-\overrightarrow{AB}.\overrightarrow{BC}=-|\overrightarrow{CA}||\overrightarrow{BC}|\) 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  | \(\overrightarrow{a}\) | = 3, | \(\overrightarrow{b}\) | = 4, | \(\overrightarrow{c}\) | = 5.
Also \(\overrightarrow{a}.(\overrightarrow{b}+\overrightarrow{c})=\overrightarrow{0}\)
\(\overrightarrow{b}.(\overrightarrow{c}+\overrightarrow{a})=\overrightarrow{0}\)
and \(\overrightarrow{c}.(\overrightarrow{a}+\overrightarrow{b})=\overrightarrow{0}\)
Since they are perpendicular
\(\Rightarrow \overrightarrow{a}.\overrightarrow{b}+\overrightarrow{a}.\overrightarrow{c}=0,\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{b}.\overrightarrow{a}=0\) and \(\overrightarrow{c}.\overrightarrow{a}+\overrightarrow{c}.\overrightarrow{b}=0\)
Adding all the above we get,
\(2(\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a})=0\)
\(\Rightarrow \overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a}=0\)..(1)
Consider \(|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}|^2=|\overrightarrow{a}|^2+|\overrightarrow{b}|^2+|\overrightarrow{c}|^2+2(\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a})\)
= 9 + 16 + 25 + 2(0) = 50 
\(\therefore |\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}|=\sqrt{50}=\sqrt{25\times 2}=5\sqrt{2}\)

Let \(\overrightarrow{a}=5\hat{i}+6\hat{j}+7\hat{k}\)
\(\overrightarrow{b}=\)7 \(\hat{i}\) -8\(\hat{j}\) +9 \(\hat{k}\)
\(\overrightarrow{c}=\)3\(\hat{i}\)+20\(\hat{j}\) +5\(\hat{k}\)
Let \(\overrightarrow{a}=s\overrightarrow{b}+t \overrightarrow{c}\)
\(\Rightarrow 5\hat { i } +6\hat { j } +7\hat { k } =s(7\hat { i } -8\hat { j } +9\hat { k } )+t(3\hat { i } +20\hat { j } +5\hat { k } )\)
\(\Rightarrow 5\hat { i } +6\hat { j } +7\hat { k } =(7s+3t)\hat { i } +(-8s+20t)\hat { j } +(9s+5t)\hat { k } \)
Equating the like components, both sides we get.
                   5 = 7s + 3t  .....(1)
                  -8s + 20t = 6 ....(2)
                   9s + 5t = 7 ......(3)

                  164s  = 82    \(\Rightarrow \quad s=\frac { 82 }{ 164 } =\frac { 1 }{ 2 } \)
Substituting  \(\\ s=\frac { 1 }{ 2 } \) in (1) we get,
\(7\left( \frac { 1 }{ 2 } \right) +3t=5\quad \Rightarrow 3t=5-\frac { 7 }{ 2 } =\frac { 10-7 }{ 2 } =\frac { 3 }{ 2 } \)
\(\Rightarrow t=\frac { 3 }{ 2\times 3 } =\frac { 1 }{ 2 } \)
Substituting \(s=\frac { 1 }{ 2 } ,t=\frac { 1 }{ 2 } \) in (3) we get,
\(9\left( \frac { 1 }{ 2 } \right) +5\left( \frac { 1 }{ 2 } \right) =7\)
\(\Rightarrow \frac { 9 }{ 2 } +\frac { 5 }{ 2 } =7\)
\(\Rightarrow \frac { 14 }{ 2 } =7\)
\(\Rightarrow\) 7 = 7 which satisfies the (3) equation.
Thus, one vector is a linear combination of other two vectors.
Hence, the given vectors are co-planar.                                 

Let  \(\overrightarrow {a}=\hat{i}\) − 2\(\hat{j}\) + 3\(\hat{k}\), \(\overrightarrow{b}=\) -2\(\hat{i}\) + 3\(\hat{j}\) - 4\(\hat{k}\), \(\overrightarrow{c}=\) -\(\hat{j}\) + 2\(\hat{k}\) .
Let \(\overrightarrow {a}=s\overrightarrow{b}+t\overrightarrow{c}\)
\(\Rightarrow \hat{i}-2\hat{j}+3\hat{k}=s(-2\hat{i}+3\hat{j}-4\hat{k})+t(-\hat{j}+2\hat{k})\)
\(\Rightarrow \hat{i}-2\hat{j}+3\hat{k}=(-2s)\hat{i}+(3s-t)\hat{j}+(-4s +2t)\hat{k}\)
Equating the like components both sides, we get
-2s = 1 ....(1)
3s - t = -2 .....(2)
-4s + 2t = 3 ......(3)
From(1), s = \(-{1\over2}\)
Substituting s = \(-{1\over2}\) in (2) we get,
3\(({-1\over2})-t=-2 \Rightarrow -{3\over2}-t=-2\)
\(-t=-2+{3\over2}\)
\(-t={-4+3\over2}={-1\over2}\)
\(t={1\over2}\)
Substituting s = \(-{1\over2}\),\(t={1\over2}\) in (3) we get,
\(-4({-1\over2})+2({1\over2})=+3\)
\(\Rightarrow 2+1=3\)
\(\Rightarrow{3=3}\)
which satisfies the (3) equation.
Thus, one vector is a linear combination of other two vectors.
Hence, the given vectors are co-planar.

Let \(5\hat{i}+6\hat{j}+7\hat{k}=s(7\hat{i}-8\hat{j}+9\hat{k})+t(3\hat{i}+20\hat{j}+5\hat{k})\)
Equating the components, we have
7s + 3t = 5
-8s + 20t = 6
9s + 5t = 7
Solving first two equations, we get, s = t = \({1\over2},\) which satisfies the third equation.
Thus one vector is a linear combination of other two vectors.
Hence the given vectors are coplanar.

Let  A, B, C be the given points and O be the point of reference or origin.
Then \(\overrightarrow{OA}=2\hat{i}+4\hat{j}+3\hat{k},\overrightarrow{OB}=4\hat{i}+\hat{j}+9\hat{k} \ and \overrightarrow{OC}=10\hat{i}-\hat{j}+6\hat{k}\)
\(\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=(4\hat{i}+\hat{j}+9\hat{k})-(2\hat{i}+4\hat{j}+3\hat{k})=2\hat{i}-3\hat{j}+6\hat{k}.\)
\(AB=|\overrightarrow{AB}|=\sqrt{2^2+(-3)^2+6^2}=\sqrt{4+9+36}=7\)
\(\overrightarrow{BC}=\overrightarrow{OC}-\overrightarrow{OB}=(10\hat{i}-\hat{j}+6\hat{k})-(4\hat{i}+\hat{j}+9\hat{k})=6\hat{i}-2\hat{j}-3\hat{k}.\)
\(BC=|\overrightarrow{BC}|=\sqrt{6^2+(-2)^2+(-3)^2}=\sqrt{36+4+9}=7\)
\(\overrightarrow{CA}=\overrightarrow{OA}-\overrightarrow{OC}=(2\hat{i}+4\hat{j}+3\hat{k})-(10\hat{i}-\hat{j}+6\hat{k})=-8\hat{i}+5\hat{j}-3\hat{k}\).
\(CA=|\overrightarrow{CA}|=\sqrt{(-8)^2+5^2+(-3)^2}=\sqrt{64+25+9}=\sqrt{98}\)
\(BC^2=49,CA^2=98,AB^2=49.\)
Clearly CA² = BC² + AB².
Therefore, the given points form a right angled triangle.

Let the position vectors of the vertices of the quadrilateral be \(\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\) and \(\overrightarrow{d}\).
Let P, Q, R, S be the mid-points of the adjacent sides of the quadrilateral.
To prove that PQRS is a parallelogram.
\(\overrightarrow{OP}={\overrightarrow{a}+\overrightarrow{b}\over2},\overrightarrow{OQ}={\overrightarrow{b}+\overrightarrow{c}\over 2},\overrightarrow{OR}={\overrightarrow{c}+\overrightarrow{d}\over 2},\overrightarrow{OS}={\overrightarrow{a}+\overrightarrow{d}\over2}\)
Now, \(\overrightarrow{PQ}=\overrightarrow{OQ}-\overrightarrow{OP}={\overrightarrow{b}+\overrightarrow{c}\over 2}-{\overrightarrow{a}+\overrightarrow{d}\over2}={\overrightarrow{b}+\overrightarrow{c}-\overrightarrow{a}-\overrightarrow{b}\over2}\)
\(={\overrightarrow{c}-\overrightarrow{a}\over2}\).....(1)
\(\overrightarrow{SR}=\overrightarrow{OR}-\overrightarrow{OS}={\overrightarrow{c}+\overrightarrow{d}\over 2}-{\overrightarrow{a}+\overrightarrow{d}\over2}={\overrightarrow{c}+\overrightarrow{d}-\overrightarrow{a}-\overrightarrow{d}\over2}\)
\(={\overrightarrow{c}-\overrightarrow{a}\over2}\).....(2)
From (1) and (2),\(\overrightarrow{PQ}=\overrightarrow{SR}\)
and \(\overrightarrow{PQ}=1(\overrightarrow{SR}) \Rightarrow \overrightarrow{PQ}||\overrightarrow{SR}\)
Thus, one pair of parrallel sides of PQRS are parallel and equal.
\(\therefore\) PQRS is a parallelogram.