Let  \(\overrightarrow{a}\)and \(\overrightarrow{b}\) be the unit vectors and \(\theta\) is the angle between \(\overrightarrow{a}\) and \(\overrightarrow{b}\).
Consider \(|\overrightarrow{a}-\overrightarrow{b}|^2=|\overrightarrow{a}|^2+|\overrightarrow{b}|^2-2(\overrightarrow{a}.\overrightarrow{b})\)      \([\because |\overrightarrow{a}|=1;|\overrightarrow{b}|=1]\)
\(=1+1-2|\overrightarrow{a}||\overrightarrow{b}|cos \theta =2+2cos \theta\)
\(=2(1+cos \theta)=2.2cos^2{\theta \over2}=4cos^2{\theta \over2}\)
\(\therefore |\overrightarrow{a}+\overrightarrow{b}|=2cos{\theta \over2}\)
\(\Rightarrow cos{\theta \over2}={1\over 2}|\overrightarrow{a}+\overrightarrow{b}|\)

Given  | \(\overrightarrow{a}\) | = 10, | \(\overrightarrow{b}\) | = 15 and \(\overrightarrow{a}\).\(\overrightarrow{b}\) = 75 \(\sqrt{2}\)
Let \(\theta\) be the angle between the vector \(\overrightarrow{a}\) and \(\overrightarrow{b}\).

\(\theta ={\pi\over 4}.\)

Let \(\overrightarrow{r}=x\hat{i}+y\overrightarrow{j}+z\hat{k}\)
\(\overrightarrow{r}.\hat{i}=(x\hat{i}+y\hat{j}+z\hat{k}).\hat{i}=x\)
\(\overrightarrow{r}.\hat{j}=(x\hat{i}+y\hat{j}+z\hat{k}).\hat{j}=y\)
\(\overrightarrow{r}.\hat{k}=(x\hat{i}+y\hat{j}+z\hat{k}).\hat{k}=z\)
\((\overrightarrow{r}.\hat{i})\hat{i}+(\overrightarrow{r}.\hat{j})\hat{j}+(\overrightarrow{r}.\hat{k})\hat{k}=x\hat{i}+y\hat{j}+z\hat{k}=\overrightarrow{r}\)
Thus \(\overrightarrow{r}=(\overrightarrow{r}.\hat{i})\hat{i}+(\overrightarrow{r}.\hat{j})\hat{j}+(\overrightarrow{r}.\hat{k})\hat{k}\) .

Let O be the origin. It is given that
\(\overrightarrow{OA}=2\overrightarrow{a}+4\overrightarrow{b}\ and\ \overrightarrow{OB}=2\overrightarrow{a}- 8\overrightarrow{b}\)
Let C and D be the points which divide the segment AB in the ratio 1 : 3 internally and externally respectively. Then
\(\overrightarrow{OC}={3\overrightarrow{OA}+\overrightarrow{OB}\over3+1}={3(2\overrightarrow{a}+4\overrightarrow{b})+(2\overrightarrow{a}+8\overrightarrow{b})\over 4}=2\overrightarrow{a}+\overrightarrow{b}.\)
\(\overrightarrow{OC}={3\overrightarrow{OA}-\overrightarrow{OB}\over3-1}={3(2\overrightarrow{a}+4\overrightarrow{b})-(2\overrightarrow{a}+8\overrightarrow{b})\over 2}=2\overrightarrow{a}+10\overrightarrow{b}.\)

Let \(\theta\) be the angle between \(\overrightarrow{a}\) and \(\overrightarrow{c}\)
\(\overrightarrow{a}-\sqrt{3}\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}\)
\(\Rightarrow |(\overrightarrow{a}+\overrightarrow{c})|=|\sqrt{3}\overrightarrow{b}|\)
\(\Rightarrow |\overrightarrow{a}|^2+|\overrightarrow{c}|^2+2|\overrightarrow{a}||\overrightarrow{c}|cos \theta=3|{}\overrightarrow{b}|^2\)
\(\Rightarrow 1+1+(2)(1)(1)cos \theta=3(1)\)
\(\Rightarrow cos \theta ={1\over2}\Rightarrow \theta={\pi\over 3}.\)

Let the position vectors of the given vector be
\(\overrightarrow { OA } =4\hat { i } +5\hat { j } +\hat { k } \)
\(\overrightarrow { OB } =-\hat { j } -\hat { k } \)
\(\overrightarrow { OC } =3\hat { i } +9\hat { j } +4\hat { k } \)
and \(\overrightarrow { OD } =-7\hat { i } +4\hat { j } +4\hat { k } \)
Let \(\overrightarrow { a } =\overrightarrow { AB } =\overrightarrow { OB } -\overrightarrow { OA } =(-\hat { j } -\hat { k } )-(4\hat { i } +5\hat { j } +\hat { k } )=-4\hat { i } -6\hat { j } -2\hat { k } \)
\(\overrightarrow { b } =\overrightarrow { AC } =\overrightarrow { OC } -\overrightarrow { OA } =(3\hat { i } +9\hat { j } +4\hat { k } )-(4\hat { i } +5\hat { j } +\hat { k } )=-\hat { i } +4\hat { j } +3\hat { k } \)
and \(\overrightarrow { c } =\overrightarrow { AD } =\overrightarrow { OD } -\overrightarrow { OA } =(-4\hat { i } +4\hat { j } +4\hat { k } )-(4\hat { i } +5\hat { j } +\hat { k } )=-8\hat { i } -\hat { j } +3\hat { k } \)
Also, let \(\overrightarrow { a } =s\overrightarrow { b } +t\overrightarrow { c } \)
\(-4\hat { i } -6\hat { j } -2\hat { k } =s(-\hat { i } +4\hat { j } +3\hat { k } )+t(-8\hat { i } -\hat { j } +3\hat { k } )\)
\(-4\hat { i } -6\hat { j } -2\hat { k } =(-s-8t)\hat { i } +(4s-t)\hat { j } +(3s+3t)\hat { k } \)
Equating the like components on both sides, we get
-4 = -s - 8t
-6 = 4s - t
-2 = 3s + 3t
(1) \(\times\) 4 \(\Rightarrow\) -16 = -4s - 32t
(2)      \(\Rightarrow\)    -6  =  4s - t
Adding -22  = -33 t         
\(\\ \Rightarrow t=\frac { -22 }{ -33 } =\frac { 2 }{ 3 } \)
Substituting \(t=\frac { 2 }{ 3 } \) in (1) we get,
\(-4=-s-8\left( \frac { 2 }{ 3 } \right) \quad \Rightarrow -4=-s-\frac { 16 }{ 3 } \)
\(\Rightarrow s=4-\frac { 16 }{ 3 } =\frac { 12-16 }{ 3 } =-\frac { 4 }{ 3 } \)
Substituting t = \(\frac { 2 }{ 3 } \) , and s = - \(\frac { 4 }{ 3 } \) in (3) we get,
\(-2=3\left( -\frac { 4 }{ 3 } \right) +3\left( \frac { 2 }{ 3 } \right) \)
\(\Rightarrow\) - 2 = -4 + 2 \(\Rightarrow\) -2 = -2 which satisfies equation (3).
Thus, one vector is the linear combination of other two vectors.
Hence, the given points are co-planar.

Let O be the origin and let \(\overrightarrow{OA}\), \(​​\overrightarrow{OB}\), and\(​​\overrightarrow{OC}\) be the vectors 2\(\hat{i}\) + 3\(\hat{j}\) − 5\(\hat{k}\), 3\(\hat{i}\) + \(\hat{j}\) − 2\(\hat{k}\) and, 6\(\hat{i}\) − 5\(\hat{j}\) + 7\(\hat{k}\) respectively. Then
\(\overrightarrow{AB}=\hat{i}-2\hat{j}+3\hat{k} \ and \ \overrightarrow{AC}=4\hat{i}-8\hat{j}+12\hat{k}\).
Thus \(\overrightarrow{AC}=4\overrightarrow{AB}\) and hence \(​​\overrightarrow{AB}\) and\(​​\overrightarrow{AC}\) are parallel. They have a common point namely A. Thus, the three points are collinear.
Alternative method
Let O be the point of reference.
Let \(\overrightarrow {OA} = 2\hat i+3\hat j-5\hat k, \) \(\overrightarrow {OB} = 3 \hat j+\hat j-2\hat k\ and\ \overrightarrow {OC} = 6\hat i-5\hat j+7\hat k \)
\(\overrightarrow {AB} = \hat i- 2\hat j+3\hat k; \overrightarrow {BC} = 3\hat i-6\hat j+9\hat k; \overrightarrow {CA} = -4\hat i+8\hat j-12 \hat k\\ |\overrightarrow {AB}| = \sqrt 14; |\overrightarrow {BC}|= \sqrt 126 = 3 \sqrt 14; |\overrightarrow {CA}|= \sqrt 224 = 4 \sqrt 4\)
Thus, AC = AB + BC.
Hence A, B, C are lying on the same line. That is, they are collinear.

Let the position vector of the vertices of the \(\triangle\)ABC be \(\overrightarrow{a},\overrightarrow{b}\)and \(\overrightarrow{c}\) respectively.
Since D is the mid-point of BC.
\(\Rightarrow \overrightarrow{OD}={\overrightarrow{b}+\overrightarrow{c}\over 2}\)
E is the mid-point of AC,
\(\Rightarrow \overrightarrow{OE}={\overrightarrow{a}+\overrightarrow{c}\over 2}\)
and F is the mid-point of AB
\( \overrightarrow{OF}={\overrightarrow{a}+\overrightarrow{b}\over 2}\)
To prove that \(\overrightarrow{AD}+\overrightarrow{BE}+\overrightarrow{CF}=\overrightarrow{o}\)

LHS=\(\overrightarrow{AD}+\overrightarrow{BE}+\overrightarrow{CF}=\overrightarrow{OD}-\overrightarrow{OA}+\overrightarrow{OE}-\overrightarrow{OB}+\overrightarrow{OF}+\overrightarrow{OC}\)
\(={\overrightarrow{b}+\overrightarrow{c}\over 2}=\overrightarrow{a}+{\overrightarrow{a}+\overrightarrow{c}\over 2}-\overrightarrow{b}\)+\({\overrightarrow{a}+\overrightarrow{b}\over 2}-\overrightarrow{c}\)
\(={\overrightarrow{b}+\overrightarrow{c}-2\overrightarrow{a}+\overrightarrow{a}+\overrightarrow{c}-2\overrightarrow{b}+\overrightarrow{a}+\overrightarrow{b}-2\overrightarrow{c}\over 2}\) \(={\overrightarrow{0}\over2}=\overrightarrow{0}=RHS\)
Hence proved.

Let ABCD be the parallelogram
Let \(\overrightarrow{AB}=\overrightarrow{a}\) and \(\overrightarrow{AC}=\overrightarrow{b}\)
In \(\triangle ABC, \overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}\)
\(\Rightarrow {b}=\overrightarrow{a}+\overrightarrow{BC}\)
\(\Rightarrow \overrightarrow{BC}=\overrightarrow{b}-\overrightarrow{a}\)
Since ABCD is a parallelogram,
\(\overrightarrow{CD}=-\overrightarrow{AB}\)
\(\overrightarrow{CD}=-\overrightarrow{a}\)
\(\overrightarrow{DA}=-\overrightarrow{BC}=-(​​​​\overrightarrow{b}-\overrightarrow{a})\)
\(\Rightarrow \overrightarrow{DA}=​​​​\overrightarrow{a}-\overrightarrow{b}\)
In \(\triangle\) BCD, \(\overrightarrow{BD}=\overrightarrow{BC}+\overrightarrow{CD}=​​​​\overrightarrow{b}-\overrightarrow{a}-\overrightarrow{a}=\overrightarrow{b}-2\overrightarrow{a}\)
Hence, the other sides of the parallelogram are \(\overrightarrow{b}-\overrightarrow{a},-\overrightarrow{a},\overrightarrow{a}-\overrightarrow{b}\) and the other diagonal is \(\overrightarrow{b}-2\overrightarrow{a}\).

 
Let O be the origin.
\(\overrightarrow{O A}=\vec{a}, \overrightarrow{O B}=\vec{b}, \overrightarrow{O C}=\vec{c}\)
Since D and E are the mid points of AB and AC
\( \therefore \overrightarrow{O D} =\frac{\overrightarrow{O A}+\overrightarrow{O B}}{2}=\frac{\vec{a}+\vec{b}}{2} ; \overrightarrow{O E}=\frac{\overrightarrow{O A}+\overrightarrow{O C}}{2}=\frac{\vec{a}+\vec{c}}{2} \)
\(\text { LHS } =\overrightarrow{B E}+\overrightarrow{D C}=\overrightarrow{O E}-\overrightarrow{O B}+\overrightarrow{O C}-\overrightarrow{O D} \)
\( =\frac{\vec{a}+\vec{c}}{2}-\vec{b}+\vec{c}-\frac{(\vec{a}+\vec{b})}{2} \)
\( =\frac{\vec{a}+\vec{c}-2 \vec{b}+2 \vec{c}-\vec{a}-\vec{b}}{2} \)
\( =\frac{3 \vec{c}-3 \vec{b}}{2}=\frac{3(\vec{c}-\vec{b})}{2}=\frac{3(\overrightarrow{O C}-\overrightarrow{O B})}{2} \)
\( =\frac{3}{2} \overrightarrow{B C}=R H S\)
Hence proved.