\(\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=\hat{i}-4\hat{j}+\hat{k}\)
Direction cosines are \({1\over \sqrt{18}},{-4\over \sqrt{18}},{1\over \sqrt{18}}\).

The direction cosines are \({x\over \sqrt{x^2+y^2+z^2}},{y \over \sqrt{x^2+y^2+z^2}},{z\over \sqrt{x^2+y^2+z^2}}\)
That is, \({2\over 7},{3\over 7},{-6\over7}.\)

80km, 60° south of west

The vector \(\overrightarrow{OQ}\) represents a displacement of 80 km, 60° south of west.

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 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.

Given Now, \(\overrightarrow { a } =\hat { i } +\hat { j } ;\overrightarrow { b } =\hat { j } +\hat { k } ;\overrightarrow { c } =\hat { i } +\hat { k } \)
\(\therefore \overrightarrow { a } -2\overrightarrow { b } +3\overrightarrow { c } =(\hat { i } +\hat { j } )-2(\hat { j } +\hat { k } )+3(\hat { i } +\hat { k } )=4\hat { i } -\hat { j } +\hat { k } \)
\(\therefore |\overrightarrow { a } -2\overrightarrow { b } +3\overrightarrow { c } |=\sqrt { { 4 }^{ 2 }+{ (-1) }^{ 2 }+{ 1 }^{ 1 } } =\sqrt { 16+1+1 } =\sqrt { 18 } =\sqrt { 9\times 2 } =3\sqrt { 2 } \)
Thus, the unit vector in the direction of \(\overrightarrow { a } -2\overrightarrow { b } +3\overrightarrow { c } \) is
\(\frac { \overrightarrow { a } -2\overrightarrow { b } +3\overrightarrow { c } }{ |\overrightarrow { a } -2\overrightarrow { b } +3\overrightarrow { c } | } =\frac { 1 }{ 3\sqrt { 2 } } (4\hat { i } -\hat { j } +\hat { k } )\)

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

\(\text { Unit vector }=\frac{\lambda \hat{i}+2 \lambda \hat{j}+2 \lambda \hat{k}}{\sqrt{\lambda^{2}+4 \lambda^{2}+4 \lambda^{2}}}\)
\(=\frac{\lambda \hat{k}+2 \lambda \hat{j}+2 \lambda \hat{k}}{\sqrt{9 \lambda^{2}}}=\frac{\lambda \hat{i}+2 \lambda \hat{j}+2 \lambda \hat{k}}{3 \lambda} \)
\(=\frac{\lambda(\hat{i}+2 \hat{j}+2 \hat{k})}{3 \lambda}=\frac{1}{3}(\hat{i}+2 \hat{j}+2 \hat{k}) \)
\(\lambda =\frac{1}{3} \)

\(2 \vec{a}=\vec{b}+\vec{c} \Rightarrow \vec{a}+\vec{a}=\vec{b}+\vec{c} \Rightarrow \vec{a}-\vec{b}=\vec{c}-\vec{a} \)
\(\overrightarrow{O A}-\overrightarrow{O B}=\overrightarrow{O C}-\overrightarrow{O A} \Rightarrow \overrightarrow{B A}=\overrightarrow{A C} \)
\(\Rightarrow \vec{a}, \vec{b}, \vec{c} \text { are collinear }\)

All angle are equal
\(\therefore \alpha=\beta=\gamma\)
\(\text { W.K.T } \cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1\)
\(\cos ^{2} \alpha+\cos ^{2} \alpha+\cos ^{2} \alpha =1 \)
\(3 \cos ^{2} \alpha =1 \)
\(\cos ^{2} \alpha =\frac{1}{3} \)
\(\cos \alpha =\pm \frac{1}{\sqrt{3}} \)
\(\alpha =\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right) \)

\(\underbrace{\overrightarrow{A B}}+ \overrightarrow{B C}+\overrightarrow{C D}+\overrightarrow{D A} \)
\(=\underbrace{\overrightarrow{A C}+\overrightarrow{C D}}+\overrightarrow{D A} \)
\(=\underbrace{\overrightarrow{A D}+\overrightarrow{D A}} \)
\(=\overrightarrow{A A}=\overrightarrow{0} . \)