We have
\(\overrightarrow{\mathrm{AB}}=(1-2) \hat{i}+(-3+1) \hat{j}+(-5-1) \hat{k}=-\hat{i}-2 \hat{j}-6 \hat{k}\)
\(\overrightarrow{\mathrm{BC}}=(3-1) \hat{i}+(-4+3) \hat{j}+(-4+5) \hat{k}=2 \hat{i}-\hat{j}+\hat{k} \)
\(\text {and}\overline{\mathrm{CA}}=(2-3) \hat{i}+(-1+4) \hat{j}+(1+4) \hat{k}=-\hat{i}+3 \hat{j}+5 \hat{k} \)
Further, note that
\(|\overrightarrow{\mathrm{AB}}|^{2}=41=6+35=|\overrightarrow{\mathrm{BC}}|^{2}+|\overrightarrow{\mathrm{CA}}|^{2}\)
Hence, the triangle is a right angled triangle.

Note that two vectors are equal if and only if their corresponding components are equal. Thus, the given vectors \(\vec{a}\) and \(\vec{b}\)  will be equal if and only if
x = 2, y = 2, z = 1

(i) Time-scalar      
(ii) volume scalar
(iii) Force-vector    
(iv) speed-scalar
(v) desity-scalar    
(vi) velocity-vector.

\((\overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } ).(\overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } )=\overset { \rightarrow }{ a } .(\overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } )+\overset { \rightarrow }{ b } .(\overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } )\)
\(=\overset { \rightarrow }{ a } .\overset { \rightarrow }{ a } +\overset { \rightarrow }{ a } .\overset { \rightarrow }{ b } +\overset { \rightarrow }{ b } .\overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } .\overset { \rightarrow }{ b } \)
\(\overset { \rightarrow }{ { |a| }^{ 2 } } +\overset { \rightarrow }{ a } .\overset { \rightarrow }{ b } +\overset { \rightarrow }{ b } .\overset { \rightarrow }{ a } +\overset { \rightarrow }{ |{ b| }^{ 2 } } .\)
\(\overset { \rightarrow }{ { |a| }^{ 2 } } +\overset { \rightarrow }{ |{ b| }^{ 2 } } +2\overset { \rightarrow }{ a } .\overset { \rightarrow }{ b } \)..(1)
When \(\overset { \rightarrow }{ a } ,\overset { \rightarrow }{ b } \) are perpendicular \(\Rightarrow \) \(\overset { \rightarrow }{ a } .\overset { \rightarrow }{ b } \) = 0
\(\therefore\) From (1),
\((\overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } ).(\overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } )=\overset { \rightarrow }{ { |a| }^{ 2 } } +\overset { \rightarrow }{ { |b| }^{ 2 } } \)
\(\Rightarrow \overset { \rightarrow }{ a } .\overset { \rightarrow }{ b } =0\Rightarrow \overset { \rightarrow }{ a } ,\overset { \rightarrow }{ b } \ \) are perpendicular.

Let \(\overset { \rightarrow }{ a } =\overset { \wedge }{ xi } +\overset { \wedge }{ xj } +\overset { \wedge }{ xk } \) 
By the question, \(\overset { \rightarrow }{ |a| } =1\) 
\(\Rightarrow \sqrt { { x }^{ 2 }+{ x }^{ 2 }+{ x }^{ 2 } } = 1\)
\(\Rightarrow \sqrt { 3x } =\pm \Rightarrow x=\pm \frac { 1 }{ \sqrt { 3 } } \)  

We have : \(\overset { \rightarrow }{ a } =\overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } \) 
Squaring \(\overset { \rightarrow }{ { |a| }^{ 2 } } = { \left| \overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } \right| }^{ 2 }\)
\( =\left( \overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } \right) .\left( \overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } \right) \) 
\(=\overset { \rightarrow }{ b } .\overset { \rightarrow }{ b } +\overset { \rightarrow }{ b } .\overset { \rightarrow }{ c } +\overset { \rightarrow }{ c } .\overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } .\overset { \rightarrow }{ c } \)
\(=\overset { \rightarrow }{ { |b| }^{ 2 } } +2\overset { \rightarrow }{ b } .\overset { \rightarrow }{ c } +\overset { \rightarrow }{ { |c| }^{ 2 } } \left[ \because \overset { \rightarrow }{ b } .\overset { \rightarrow }{ c } =\overset { \rightarrow }{ c } .\overset { \rightarrow }{ b } \right] \)
\(=\overset { \rightarrow }{ { |b| }^{ 2 } } +\overset { \rightarrow }{ { |c| }^{ 2 } } +2\overset { \rightarrow }{ |b| } \overset { \rightarrow }{ |c| } cos\theta \)
Where '\(\theta \)' is the  angle between \(\overset { \rightarrow }{ b } \ and\ \overset { \rightarrow }{ c } \) 
When \(\theta \) =0\(°\) , then \(\overset { \rightarrow }{ { |a| }^{ 2 } } \) 
\(=\overset { \rightarrow }{ { |b| }^{ 2 } } +\overset { \rightarrow }{ { |c| }^{ 2 } } +2\overset { \rightarrow }{ |b| } \overset { \rightarrow }{ |c| } \)
\(=\)\(\overset { \rightarrow }{ |a| } =\overset { \rightarrow }{ |b } |+\overset { \rightarrow }{ |c| } \) 
When \(\theta\) \(\neq \) 0, then \(\overset { \rightarrow }{ |a| } \neq \overset { \rightarrow }{ |b } |+\overset { \rightarrow }{ |c| } \) 

When \(\overset { \rightarrow }{ a } =\overset { \rightarrow }{ 0 } \) and \(\overset { \rightarrow }{ |a| } =\overset { \rightarrow }{ 0 } \) then
\(\overset { \rightarrow }{ |a| } =\overset { \rightarrow }{ 0 } \ \overset { \rightarrow }{ b } =\overset { \rightarrow }{ 0 } \quad \overset { \rightarrow }{ a } *\overset { \rightarrow }{ b } =\overset { \rightarrow }{ |a| } \overset { \rightarrow }{ |b| } \quad sin\theta \overset { \wedge }{ n } \)
Whenever '\(\theta \) is the angle betwwn \(\vec{a} \text { and } \vec{b}=(0) \overrightarrow{|b|} \sin \theta_{n}^{\wedge}\) 
Similarly When \(\overset { \rightarrow }{ b } =\overset { \rightarrow }{ 0 } \), then \(\overset { \rightarrow }{ a } \times \overset { \rightarrow }{ b } =0\) 
Conversly: Let  \(\\ \\ \overset { \rightarrow }{ a } ={ a }_{ 1 }\overset { \wedge }{ i } +{ a }_{ 2 }\overset { \wedge }{ j } +{ a }_{ 3 }\overset { \wedge }{ k }\)
\( and\ \overset { \rightarrow }{ b } ={ \mu }_{ a1 }\overset { \wedge }{ i } +{ \mu }_{ a2 }\overset { \wedge }{ j } +{ \mu }_{ a3 }\overset { \wedge }{ k } \)
Clearly \(\overset { \rightarrow }{ a } \) and \(\overset { \rightarrow }{ b } \) are parallel
\(\Rightarrow \theta =0\)
When \(\overset { \rightarrow }{ |a } |\neq 0\ and\ \overset { \rightarrow }{ |b| } \neq 0\)
But \(\overset { \rightarrow }{ a } x\overset { \rightarrow }{ b } =\overset { \rightarrow }{ |a } |\quad \overset { \rightarrow }{ |b| } \ sin\theta \overset { \wedge }{ n } =0\) 
Hence, \(\overset { \rightarrow }{ a } x\overset { \rightarrow }{ b } =0\ even\ \overset { \rightarrow }{ a } \neq 0,\ and\ \overset { \rightarrow }{ b } \neq 0.\) 

Given: \(\overset { \rightarrow }{ a } .\overset { \rightarrow }{ b } =0\) and \(\overset { \rightarrow }{ a } *\overset { \rightarrow }{ b } =0\)  
\(\Rightarrow \left( \overset { \rightarrow }{ a } =\overset { \rightarrow }{ 0 } \ or\ \overset { \rightarrow }{ b } =\overset { \rightarrow }{ 0 } \ or\ \overset { \rightarrow }{ a } \bot \overset { \rightarrow }{ b } \right) \)
\(and\left( \overset { \rightarrow }{ a } =\overset { \rightarrow }{ 0 } \ or\ \overset { \rightarrow }{ b } =\overset { \rightarrow }{ 0 } \ or\ \overset { \rightarrow }{ a } =||\overset { \rightarrow }{ b } \right)\)
\(\Rightarrow either\overset { \rightarrow }{ a } =\overset { \rightarrow }{ 0 } \quad or\quad \overset { \rightarrow }{ b } =\overset { \rightarrow }{ 0 } \)
\(\left[ \because \overset { \rightarrow }{ a } \bot \overset { \rightarrow }{ b } \quad and\quad \overset { \rightarrow }{ a } ||\overset { \rightarrow }{ b } \ are\ not\ valid\ at\ the\ same\ time \right] \)

We have \(\left( \overset { \rightarrow }{ x } -\overset { \rightarrow }{ a } \right) .\left( \overset { \rightarrow }{ x } +\overset { \rightarrow }{ a } \right) =12\)
\(\Rightarrow \) \(\overset { \rightarrow 2 }{ x } -\overset { \rightarrow 2 }{ a } =12\Rightarrow \overset { \rightarrow 2 }{ x } -\overset { \rightarrow }{ { |a| }^{ 2 } } =12\)
\(\Rightarrow \) \(\overset { \rightarrow 2 }{ x } -{ 1 }^{ 2 }=12\Rightarrow \overset { \rightarrow 2 }{ { |x| }^{ 2 } } =13\)
hence \(\overset { \rightarrow }{ |x| } =\sqrt { 18 } \)

If \(\theta \) be the angle between \(\overset { \rightarrow }{ a } \) and \(\overset { \rightarrow }{ b } \), then
cos \(\theta \) = \(\frac { \overset { \rightarrow }{ a } ,\overset { ​​​​\rightarrow }{ b } }{ |\overset { \rightarrow }{ a } ||\overset { \rightarrow }{ b } | } =\frac { \sqrt { 6 } }{ \sqrt { 3(2) } } =\frac { \left( \sqrt { 3 } \right) \left( \sqrt { 2 } \right) }{ \left( \sqrt { 3 } \right) \left( 2 \right) } \)
= \(\frac { 1 }{ \sqrt { 2 } } =cos\frac { \pi }{ 4 } \)
Hence, '\(\theta \)' = \(\frac { \pi }{ 4 } \) 

\(\text { Let } \vec{a}=\hat{i}+2 \hat{j}+3 \hat{k} \)
\(\therefore|\vec{a}|=\sqrt{1^{2}+2^{2}+3^{2}}=\sqrt{1+4+9}=\sqrt{14} \)
\(\text { Hence, the direction cosines of }\vec{a} \text { are }\left(\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}\right)\) 

\(\text { Let } \vec{a}=5 \hat{i}-\hat{j}+2 \hat{k} .\)
\(\therefore|\vec{a}|=\sqrt{5^{2}+(-1)^{2}+2^{2}}=\sqrt{25+1+4}=\sqrt{30} \)
\(\therefore \hat{a}=\frac{\vec{a}}{|\vec{a}|}=\frac{5 \hat{i}-\hat{j}+2 \hat{k}}{\sqrt{30}} \)
Hence, the vector in the direction of vector  \(5 \hat{i}-\hat{j}+2 \hat{k}\) which has magnitude 8 units is given by  
\(8 \hat{a}=8\left(\frac{5 \hat{i}-\hat{j}+2 \hat{k}}{\sqrt{30}}\right)=\frac{40}{\sqrt{30}} \hat{i}-\frac{8}{\sqrt{30}} \hat{j}+\frac{16}{\sqrt{30}} \hat{k} \)
\(=8\left(\frac{5 \vec{i}-\vec{j}+2 \vec{k}}{\sqrt{30}}\right) \)
\(=\frac{40}{\sqrt{30}} \vec{i}-\frac{8}{\sqrt{30}} \vec{j}+\frac{16}{\sqrt{30}} \vec{k} \)

The given points are  P(1, 2, 3) and Q(4, 5, 6)
\(\therefore \overrightarrow{\mathrm{PQ}}=(4-1) \hat{i}+(5-2) \hat{j}+(6-3) \hat{k}=3 \hat{i}+3 \hat{j}+3 \hat{k} \)
\(|\overrightarrow{\mathrm{PQ}}|=\sqrt{3^{2}+3^{2}+3^{2}}=\sqrt{9+9+9}=\sqrt{27}=3 \sqrt{3} \)
Hence, the unit vector in the direction of \( \overrightarrow{\mathrm{PQ}} \ {\text {is }}\)
\(\frac{\overrightarrow{\mathrm{PQ}}}{|\overrightarrow{\mathrm{PQ}}|}=\frac{3 \hat{i}+3 \hat{j}+3 \hat{k}}{3 \sqrt{3}}=\frac{1}{\sqrt{3}} \hat{i}+\frac{1}{\sqrt{3}} \hat{j}+\frac{1}{\sqrt{3}} \hat{k}\) 

The two vectors \( 2 \hat{i}+3 \hat{j} \text { and } x \hat{i}+y \hat{j} \)will be equal if their corresponding components are equal.
Hence, the required values of  x and y are 2 and 3 respectively. \(\)

\(\text { Consider } \vec{a}=(\hat{i}-2 \hat{j}+3 \hat{k}) \text { and } \vec{b}=(2 \hat{i}+\hat{j}-3 \hat{k}) \text { . }\)
\(\text { It can be observed that }|\vec{a}|=\sqrt{1^{2}+(-2)^{2}+3^{2}}=\sqrt{1+4+9}=\sqrt{14} \text { and }\)
\(|\vec{b}|=\sqrt{2^{2}+1^{2}+(-3)^{2}}=\sqrt{4+1+9}=\sqrt{14}\)
\(\text {Hence, } {\vec{a} } \text { and } \vec{b} \)are two different vectors having the same magnitude. The vectors are different because they have different directions.