Let \(\theta\) be the angle between \(\overrightarrow{a} \) and \( \overrightarrow{b}\) 
\(\overrightarrow{a}.\overrightarrow{b}=(2\hat{i}+\hat{j}+3\hat{k}).\)\((4\hat{i}-2\hat{j}+2\hat{k})\) = 8 - 2 + 6 = 12
\(|\overrightarrow{a}|=|2\hat{i}+\hat{j}+3\hat{k}|=\sqrt{14};\) \(|\overrightarrow{b}|=|4\hat{i}-2\hat{j}+2\hat{k}|=\sqrt{24}\)
\(cos \theta ={\overrightarrow{a}.\overrightarrow{b}\over |\overrightarrow{a}||\overrightarrow{b}|}={12\over \sqrt{14}\sqrt{24}}=\sqrt{3\over 7}\)
\(\overrightarrow{a}\times\overrightarrow{b} =\)\(\begin{vmatrix} \hat { i } & \hat { j } & \hat { k } \\ 2 & 1 & 3 \\ 4 & -2 & 2 \end{vmatrix}\) \(=\hat{i}(2+6)-\hat{j}(4-12)+\hat{k}(-4-4)=8\hat{i}+8\hat{j}-8\hat{k}\)
\(|\overrightarrow{a}\times\overrightarrow{b} |=|8\hat{i}+8\hat{j}-8\hat{k}|=8\sqrt{3}\)
\(sin \theta={|\overrightarrow{a}\times \overrightarrow{b}|\over |\overrightarrow{a}|| \overrightarrow{b}|}={8\sqrt{3}\over \sqrt{14}\sqrt{24}}={4\sqrt{3}\over \sqrt{7}\sqrt{12}}={2\over \sqrt{7}}\).

Trivially they form a triangle. It is enough to prove one angle is \({\pi\over2}\). So find the sides of the triangle.
Let O be the point of reference and A, B, C be (4, - 3, 1), (2, - 4, 5) and (1, - 1, 0) respectively.
\(\overrightarrow{OA}=4\hat{i}-3\hat{j}+\hat{k}\),\(\overrightarrow{OB}=2\hat{i}-4\hat{j}+5\hat{k}\), \(\overrightarrow{OC}=\hat{i}-\hat{j}\)
Now, \(\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=-2\hat{i}-\hat{j}+4\hat{k}\)
Similarly, \(\overrightarrow{BC}=-\hat{i}+3\hat{j}-5\hat{k};\overrightarrow{CA}=3\hat{i}-2\hat{j}+\hat{k}\)
Clearly, \(\overrightarrow{AB}.\overrightarrow{CA}=0\)
Thus one angle is \({\pi\over2}\). Hence they form a right angled triangle.

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.

Given \(\overrightarrow{a}=3\hat{i}+2\hat{j}+9\hat{k} \), \(\overrightarrow{b}=\overrightarrow{i}+\lambda \overrightarrow{j}+3\overrightarrow{k}\)
Given \(\overrightarrow{a}||\overrightarrow{b}\)
\(\therefore \overrightarrow{a}=\) (some scalar) \(\overrightarrow{b}\)
\(\Rightarrow \overrightarrow{a}= 3\hat{i}+2\hat{j}+9\hat{k}=3(\hat{i}+{2\over3}\hat{j}+3\hat{k})\)
\(\overrightarrow{a}=3(\overrightarrow{b})\)
\(\overrightarrow{b}=\hat{i}+{2\over3}\hat{j}+3\hat{k}\)
Comparing this\(\hat{i}+\lambda \hat{j}+3\hat{k}\) with we get 
\(\lambda ={2\over3}\)

Given \(|\overrightarrow{a}|=5,|\overrightarrow{b}|=6,|\overrightarrow{c}|=7\) and \(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} =\overrightarrow{0}\) 
\(|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}| =|\overrightarrow{a}|^2+|\overrightarrow{b}|^2+|\overrightarrow{c}|^2+2(\overrightarrow{a}.\overrightarrow{b})+2(\overrightarrow{b}.\overrightarrow{c})+2(\overrightarrow{c}.\overrightarrow{a})\)
\(=25+36+49+2(\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a})=\overrightarrow{0}\)
\(\Rightarrow -110=2(\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a})\)
\(\Rightarrow {-110\over 2}=\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a}\)
\(\Rightarrow \overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a}=-55\)

Given \(\overrightarrow{a}+2\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}\)
and \(|\overrightarrow{a}|=3,|\overrightarrow{b}|=4\) and \(|\overrightarrow{c}|=7\)
Let \(\theta\) be the angle between \(\overrightarrow{a}\) and \(\overrightarrow{b}\).
\(\Rightarrow \overrightarrow{a} +2\overrightarrow{b}=-\overrightarrow{c}\)
\(\Rightarrow |\overrightarrow{a} +2\overrightarrow{b}|^2=|-\overrightarrow{c}|^2\)
\(\Rightarrow |\overrightarrow{a}|^2 +4|\overrightarrow{b}|^2+4(\overrightarrow{a}.\overrightarrow{b})=|\overrightarrow{c}|^2\)
\(\Rightarrow 9+4(16)+4(\overrightarrow{a}.\overrightarrow{b})=49\)
\(9+64+4(\overrightarrow{a}.\overrightarrow{b})=49\)
\(\Rightarrow 73+4(\overrightarrow{a}.\overrightarrow{b})=49\)
\(\Rightarrow 4(\overrightarrow{a}.\overrightarrow{b})=49-73\)
\(\Rightarrow 4|\overrightarrow{a}||\overrightarrow{b}|cos \theta=-24\)
\(\Rightarrow 4(3)(4)cos \theta=-24\)

\(\Rightarrow cos \theta =cos(\pi-{\pi\over 3})=cos{2\pi\over 3}\)
\(\Rightarrow \theta ={2\pi\over 3}\)

Let \(\overrightarrow{a}=5\hat{i}+3\hat{j}+4\hat{k},\) and \(\overrightarrow{b}=6\hat{i}-8\hat{j}-\hat{k}\) .
Let \(\theta\) be the angle between them.
\(cos \theta ={\overrightarrow{a}.\overrightarrow{b}\over|\overrightarrow{a}||\overrightarrow{b}|}={30-24-4\over \sqrt{50}\sqrt{101}}={\sqrt{2}\over 5\sqrt{101}}\Rightarrow \theta =cos^{-1}\left[ \sqrt{2}\over 5\sqrt{101} \right] \).

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 \(\overrightarrow{r}=(\overrightarrow{r}.\hat{i})\hat{i}+(\overrightarrow{r}.\hat{j})\hat{j}+(\overrightarrow{r}.\hat{k})\hat{k}\) .

\(|\overrightarrow{a}+\overrightarrow{b}|=|\overrightarrow{a}-\overrightarrow{b}|\)
\(|\overrightarrow{a}+\overrightarrow{b}|^2=|\overrightarrow{a}-\overrightarrow{b}|^2\)
\(|\overrightarrow{a}|^2+|\overrightarrow{b}|^2+2\overrightarrow{a}.\overrightarrow{b}=|\overrightarrow{a}|^2+|\overrightarrow{b}|^2-2\overrightarrow{a}.\overrightarrow{b}\)
\(\Rightarrow 4\overrightarrow{a}.\overrightarrow{b}=0\)
\(\overrightarrow{a}.\overrightarrow{b}=0\)
Hence \(\overrightarrow{a}\) and \(\overrightarrow{b}\)are perpendicular.