\(\text { As } \frac{1}{k}, \frac{1}{k}, \frac{1}{k} \)are direction cosines of a line
\(\therefore \frac{1}{k^{2}}+\frac{1}{k^{2}}+\frac{1}{k^{2}}=1 \Rightarrow \frac{3}{k^{2}}=1 \)
\(\Rightarrow k^{2}=3 \)
\(\Rightarrow k=\pm \sqrt{3}\)

\(\frac{x-1}{2}=\frac{y+1}{5}=\frac{z-3}{-5}\)

\(\frac{x-5}{2}=\frac{y+2}{-1}=\frac{z-4}{3}\)

\(=-\frac{2}{3}, \frac{2}{3}, \frac{-1}{3}\)

We know \(\left| \overrightarrow a \times \overset\rightarrow b \right| =\left| \overrightarrow { a } \right| \left| \overrightarrow { b} \right| \sin\theta\)
\(\Rightarrow \sin \theta=\frac{\left| \overrightarrow { a } \times \overrightarrow { b } \right| }{\left| \overrightarrow {a } \right| \left| \overrightarrow {b } \right| }=\frac{12}{8\times3}=\frac{1}{2}\)
\(\therefore \theta=\frac{\pi}{6}\)

For vectors to be collinear \(\frac{2}{a}=\frac{-3}{6}=\frac{4}{-8}\)
a = -4

\(\left|\begin{array}{rrr} \hat{i} & \hat{j} & \hat{k} \\ 2 & 6 & 14 \\ 1 & -\lambda & 7 \end{array}\right|=\overrightarrow{0}\)
\( \Rightarrow \hat{i}(42+14 \lambda)-\hat{j}(14-14) +\hat{k}(-2 \lambda-6)=\overrightarrow{0} \Rightarrow \lambda=-3 \)

\(\Rightarrow \overrightarrow{x^{2}}-\overrightarrow{p^{2}}=80 \Rightarrow|\vec{x}|^{2}=81 \Rightarrow|\vec{x}|=9\)

\(\text { Projection of } \vec{a} \text { on } \vec{b}=\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}=\frac{8}{7}\)

Point through which line passes is (2, 1, - 4) and dr's: 1, 1,-1.
\(\therefore\) Cartesian equation of line is \(\frac{z-2}{1}=\frac{y-1}{-1}=\frac{z+4}{-1}\)

The area of a parallelogram with \( \vec{a} \text { and } \vec{b}\) as its adjacent sides is given by \(|\vec{a} \times \vec{b}| \text { . }\)
\(\vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 4 \\ 1 & -1 & 1 \end{array}\right|=5 \hat{i}+\hat{j}-4 \hat{k}\)
\(\text {Therefore } \ |\vec{a} \times \vec{b}|=\sqrt{25+1+16}=\sqrt{42}\)
and hence, the required area is \(\sqrt{42} \text { . }\)

The given lines are:
\(\frac { x-1 }{ -3 } =\frac { y-2 }{ 2k } =\frac { z-3 }{ 2 } \)...(i)
\(\text {and} \ \frac { x-1 }{ 3k } =\frac { y-1 }{ 1 } =\frac { z-6 }{ -5 } \).....(ii)
The direction-ratios of line(i) are <-3, 2k, 2>
The direction-ratios of line(ii) are <3k, 1, -5>.
The lines (1) and (2) are perpendicular
if a₁a₂ + b₁b₂ + c₁c₂ = 0
\(\therefore-3(3 k)+2 k \times 1+2(-5)=0 \)
\(\Rightarrow-9 k+2 k-10=0 \)
\(\Rightarrow 7 k=-10 \)
\(\Rightarrow k=\frac{-10}{7} \)
Therefore, for \( k=-\frac{10}{7} \) the given lines are perpendicular to each other.

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, '\(\theta \)' = \(\frac { \pi }{ 4 } \) 

\(\left| \frac { -16-36-64 }{ \sqrt { 16+36+64 } } \right| =\left| \frac { -116 }{ \sqrt { 116 } } \right| =2\sqrt { 29 } \) units

Let line through point (2, - 1, 3) is
\(\vec{r}=(2 \hat{i}-\hat{j}+3 \hat{k})+\lambda^{\prime}(a \hat{i}+b \hat{j}+c \hat{k})\)
\(\text { If line }(i) \text { is perpendicular to lines }\)
\(\vec{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda(2 \hat{i}-2 \hat{j}+\hat{k}) \)
\(\text {and } \vec{r}=(2 \hat{i}-\hat{j}-3 \hat{k})+\mu(\hat{i}+2 \hat{j}+2 \hat{k}) \)
\(\text { then }(a \hat{i}+b \hat{j}+c \hat{k}) \cdot(2 \hat{i}-2 \hat{j}+\hat{k})=0\)
\(\Rightarrow 2 a-2 b+c=0\)
\(\text { and }(a \hat{i}+b \hat{j}+c \hat{k}) \cdot(\hat{i}+2 \hat{j}+2 \hat{k})=0 \Rightarrow a+2 b+2 c=0\)
\(\Rightarrow \frac{a}{4-2}=\frac{-b}{4-1}=\frac{c}{4+2}, \text { i.e. } \frac{a}{6}=\frac{b}{-3}=\frac{c}{6} \)
\(\Rightarrow a: b: c \text { is }-6:-3: 6 \text { or } 2: 1:-2 \)
 line is \(\vec { r } =(2\hat { i } +\hat { j } -3\hat { k } )+\lambda '(2\hat { i } -\hat { j } +2\hat { k } )\)

\(\vec{r} =2 \vec{a}-\vec{b}+3 \vec{c} \)
\(=2 \hat{i}+2 \hat{j}+2 \hat{k}-4 \hat{i}+2 \hat{j}-3 \hat{k}+3 \hat{i}-6 \hat{j}+3 \hat{k} \)
\(\Rightarrow \vec{r}=\hat{i}-2 \hat{j}+2 \hat{k} \)
Vector of magnitude 6 units and parallel to
\((2 \hat a-\hat b+3\hat c) \text { is } 6 \hat{r}\)
\(\text { Vector }=6\left(\frac{\hat{i}-2 \hat{j}+2 \hat{k}}{\sqrt{1+4+4}}\right)=2 \hat{i}-4 \hat{j}+4 \hat{k}\)

Let  \(\hat{a}=\hat{i}+\hat{j}+\hat{k} \)
\(\hat{b}=2 \hat{i}+4 \hat{j}-5 \hat{k} \) 
and \(\hat{c}=\lambda \hat{i}+2 \hat{j}+3 \hat{k}\) 
now the unit vector along \(\overset { \wedge }{ b } +\overset { \wedge }{ c } \) 
\(=\frac { (\lambda +2)\overset { \wedge }{ i } +6\overset { \wedge }{ j } -2\overset { \wedge }{ k } }{ \sqrt { { (\lambda +2) }^{ 2 } } +36+4 } \) 
By the question
(\(\hat { i } +\hat { j } +\hat { k } \)).\(=\frac { (\lambda +2)\overset { \wedge }{ i } +6\overset { \wedge }{ j } -2\overset { \wedge }{ k } }{ \sqrt { { (\lambda +2) }^{ 2 } } +40 } \) =1
\(=\frac { 1 }{ \sqrt { { (\lambda +2) }^{ 2 } } +40 } (\lambda +2+6)= 6\)
\(\Rightarrow \lambda +6=\sqrt { { (\lambda +2) }^{ 2 }+40 } \)
Squaring, \({ \lambda }^{ 2 }+12\lambda +36\) 
\(={ \lambda }^{ 2 }+4\lambda +4+40\)
\(\Rightarrow 12\lambda +36=4\lambda +44\)
\(\Rightarrow 8\lambda =8\)
\(Hence\ \lambda =1\)

As 3 x \(\frac{k^2}{9}\) = 1 ⇒ k 土\(\sqrt3\)

As \({ cos }^{ 2 }\alpha +{ cos }^{ 2 }\beta +{ cos }^{ 2 }\gamma =1\)
\(\Rightarrow \frac { 1+cos2\alpha }{ 2 } +\frac { 1+cos2\beta }{ 2 } +\frac { 1+cos2\gamma }{ 2 } =1\)
= cos 2α + cos 2β + cos 2γ = 1

As direction ratios are 1, 1, 1 and direction cosines \(\frac { 1 }{ \sqrt { 3 } } ,\frac { 1 }{ \sqrt { 3 } } ,\frac { 1 }{ \sqrt { 3 } } \)
⇒ cos α = cos β = cos γ
⇒ α = β = γ

(i) (c): Position vector of \(\overrightarrow{A B} \) 
\(=(2-1) \hat{i} \dot{+}(1-1) \hat{j}+(3-1) \hat{k}=\hat{i}+2 \hat{k}\)
(ii) (b): Position vector of  \(\overrightarrow{A C} \)
\(=(3-1) \hat{i}+(2-1) \hat{j}+(2-1) \hat{k}=2 \hat{i}+\hat{j}+\hat{k}\)
(iii) (d): Position vector of \(\overrightarrow{AD} \)
\(=(3-1) \hat{i}+(3-1) \hat{j}+(4-1) \hat{k}=2 \hat{i}+2 \hat{j}+3 \hat{k}\)
(iv) (b): Area of \(\Delta A B C\) = \(\frac{1}{2}|\overrightarrow{A B} \times \overrightarrow{A C}|\)
\(\overrightarrow{A B} \times \overrightarrow{A C}=\left|\begin{array}{lll} \hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & 2 \\ 2 & 1 & 1 \end{array}\right|=\hat{i}(0-2)-\hat{j}(1-4)+\hat{k}(1-0)\)
\(=-2 \hat{i}+3 \hat{j}+\hat{k}\)
\( \Rightarrow |\overrightarrow{A B} \times \overrightarrow{A C}| =\sqrt{(-2)^{2}+3^{2}+1^{2}} \)
\(=\sqrt{4+9+1}=\sqrt{14}\)
Area of \(\Delta A B C\) \(=\frac{1}{2} \sqrt{14} \text { sq. units }\)
(v) (a): Unit vector along \(\overrightarrow{A D}=\frac{\overrightarrow{A D}}{|\overrightarrow{A D}|}\)
\(=\frac{2 \hat{i}+2 \hat{j}+3 k}{\sqrt{2^{2}+2^{2}+3^{2}}}=\frac{2 \hat{i}+2 \hat{j}+3 \hat{k}}{\sqrt{4+4+9}}=\frac{1}{\sqrt{17}}(2 \hat{i}+2 \hat{j}+3 \hat{k})\)