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

As let P divides the join of (3, 5, -1) and (6, 3, -2) in the ratio k : 1
\(\therefore \frac { 3k+5 }{ k+1 } =2\)
\(\Rightarrow 3k+5=2k+2\Rightarrow k=-3\)
∴ x-coordinate is
\(\frac { 6k+3 }{ k+1 } =\frac { -18+3 }{ -3+1 } =\frac { 15 }{ 2 } \)

As \(\left| \begin{matrix} 1 & 1 & -3 \\ 2 & 1 & -\lambda \\ 5 & 2 & 3 \end{matrix} \right| =0\)
⇒ 1(3 + 2λ) - 1(6 + 5λ) - 3(-1) = 0
⇒ 3 + 2λ - 6 - 5λ + 3 = 0
⇒ 3λ = 0
⇒ λ = 0

As \({ \left| \vec { a } \times \vec { b } \right| }^{ 2 }+({ \vec { a } .\vec { b } ) }^{ 2 }=|{ \vec { a } | }^{ 2 }|{ \vec { b } | }^{ 2 }\)
\(\Rightarrow { \left| \vec { a } \times \vec { b } \right| }^{ 2 }\) = 64 x 9 - 144 x 3
= 576 - 432 = 144
\(\Rightarrow { \left| \vec { a } \times \vec { b } \right| }^{ 2 }\) = 12

As area of parallelogram
= \(\left| \begin{matrix} \widehat { i } & \widehat { j } & \widehat { k } \\ 2 & 1 & -2 \\ 3 & 1 & -1 \end{matrix} \right| \)
= \(\left| \widehat { i } -4\widehat { j } -\widehat { k } \right| \)
= \(\sqrt { 1+16+1 } \)
= \(3\sqrt { 2 } \) sq units

As \({ \left| \sqrt { 3 } \vec { a } -\vec { b } \right| }^{ 2 }=({ \sqrt { 3 } \vec { a } -\vec { b } ) }^{ 2 }\)
\(=3\vec { { a }^{ 2 } } +\vec { { b }^{ 2 } } -2\sqrt { 3 } \vec { a } .\vec { b } \)
\(1=3+1-2\sqrt { 3 } \vec { a } .\vec { b } \)
\(\Rightarrow \vec { a } .\vec { b } =\frac { \sqrt { 3 } }{ 2 } \)
\(\therefore cos\theta =\frac { \vec { a } .\vec { b } }{ |\vec { b } ||\vec { b } | } =\frac { \sqrt { 3 } }{ 2 } \)\(\Rightarrow \theta ={ 30 }^{ 0 }\)

As vector = 15\(\left( \frac { \widehat { i } -2\widehat { j } +2\widehat { k } }{ \sqrt { 1+4+4 } } \right) \)
= \(5\widehat { i } -10\widehat { j } +10\widehat { k } \)

As position vector = \(\frac { 2(\vec { a } +\vec { b } )+1(2\vec { a } -\vec { b } ) }{ 1+2 } \) = \(\frac { 4\vec { a } +\vec { b } }{ 3 } \)

(i) (b) : The given differential equation can be written as \(\frac{d y}{d x}+\frac{\cos x}{1+\sin x} y=\frac{-x}{1+\sin x}\) 
Compare it with \(\frac{d y}{d x}+P y=Q\) ,we get
\(P=\frac{\cos x}{1+\sin x}\) and \(Q=\frac{-x}{1+\sin x}\) 
(ii) (c) : \(\text { I.F. }=e^{\int P d x}=e^{\int \frac{\cos x}{1+\sin x} d x}\) 
Put \(1+\sin x=t \Rightarrow \cos x d x=d t\) 
\(\therefore \quad \text { I.F. }=e^{\int \frac{1}{t} d t}=e^{\log t}=t=1+\sin x\) 
(iii) (d) : Solution of given differential equation is given by \(y \cdot(\mathrm{I} . \mathrm{F} .)=\int Q(\mathrm{I.F.}) d x+c\) 
\(\Rightarrow y(1+\sin x)=\int \frac{-x}{1+\sin x} \cdot(1+\sin x) d x+c\) 
\(\Rightarrow \quad y(1+\sin x)=\frac{-x^{2}}{2}+c\) 
(iv) (a) : We have, y(0) = 1 i.e., x = 0, y = 1
\(\therefore \quad 1(1+\sin 0)=c \Rightarrow c=1\) 
\(\therefore \quad y(1+\sin x)=\frac{-x^{2}}{2}+1=\frac{2-x^{2}}{2}\) 
\(\therefore \quad y=\frac{2-x^{2}}{2(1+\sin x)}\) 
(v) (b) : We have, \(y=\frac{2-x^{2}}{2(1+\sin x)}\)
\(\therefore \quad y\left(\frac{\pi}{2}\right)=\frac{2-\left(\frac{\pi}{2}\right)^{2}}{2\left(1+\sin \frac{\pi}{2}\right)}=\frac{2-\frac{\pi^{2}}{4}}{4}=\frac{8-\pi^{2}}{16}\)

(i) (b) : Given, T is the temperature of the body at any time t. Then, by Newton's law of cooling, we get \(\frac{d T}{d t}=k(T-50)\), where k is the constant of proportionality
(ii) (c) : From given information, we have
At 8 p.m. temperature is 70°F
\(\therefore\) At t = 0, T = 70°F
(iii) (b) : From given information, we have
At 10 p.m., temperature is 60°F.
\(\therefore\) At t = 2, T = 60° F
(iv) (c) : \(\frac{d T}{d t}=k(T-50) \Rightarrow \frac{d T}{T-50}=k d t\) 
On integrating both sides, we get
\(\log |T-50|=k t+\log C \Rightarrow T-50=C e^{\wedge t}\) 
Clearly, for t = 0, \(T=70^{\circ} \Rightarrow C=20\) 
Thus, T - 50 = 20e^kt
For \(t=2, T=60^{\circ} \Rightarrow 10=20 e^{2 k}\) 
\(\Rightarrow 2 k=\log \left(\frac{1}{2}\right) \Rightarrow k=\frac{1}{2} \log \left(\frac{1}{2}\right)\) 
Hence, \(T=50+20\left(\frac{1}{2}\right)^{\frac{t}{2}}\) 
(v) (b) : We have, \(T=50+20\left(\frac{1}{2}\right)^{\frac{2}{2}}\) 

(i) (b): The line along which motorcycle A is running, \(\vec{r}=\lambda(\hat{i}+2 \hat{j}-\hat{k})\) is which can be rewritten as \((x \hat{i}+y \hat{j}+z \hat{k})=\lambda \hat{i}+2 \lambda \hat{j}-\lambda \hat{k}\)
\(\Rightarrow x=\lambda, y=2 \lambda, z=-\lambda \Rightarrow \frac{x}{1}=\lambda, \frac{y}{2}=\lambda, \frac{z}{-1}=\lambda\)
Thus, the required cartesian equation is \(\frac{x}{1}=\frac{y}{2}=\frac{z}{-1}\)
(ii) (d): Clearly, D.R:s of the required line are <  1, 2, -1  >
∴ D.Cs are \( <\frac{1}{\sqrt{1^{2}+2^{2}+(-1)^{2}}}, \frac{2}{\sqrt{1^{2}+2^{2}+(-1)^{2}}}, \frac{-1}{\sqrt{1^{2}+2^{2}+(-1)^{2}}}> \)
\(\text { i.e., }<\frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}}, \frac{-1}{\sqrt{6}}>\)
(iii) (d): The line along which motorcycle B is running, is \(\vec{r}=(3 \hat{i}+3 \hat{j})+\mu(2 \hat{i}+\hat{j}+\hat{k})\), which is parallel to the vector \(2 \hat{i}+\hat{j}+\hat{k}\).
∴ D.R.'s of the required line are < 2, 1, 1 >.
(iv) (d): Here, \(\vec{a}_{1}=0 \hat{i}+0 \hat{j}+0 \hat{k}, \vec{a}_{2}=3 \hat{i}+3 \hat{j}, \vec{b}_{1}=\hat{i}+2 \hat{j}-\hat{k} \vec{b}_{2}=2 \hat{i}+\hat{j}+\hat{k}\)
\(\therefore \vec{a}_{2}-\vec{a}_{1}=3 \hat{i}+3 \hat{j}\)
and  \(\vec{b}_{1} \times \vec{b}_{2}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -1 \\ 2 & 1 & 1 \end{array}\right|=3 \hat{i}-3 \hat{j}-3 \hat{k}\) 
Now, \(\left(\vec{a}_{2}-\vec{a}_{1}\right) \cdot\left(\vec{b}_{1} \times \vec{b}_{2}\right)=(3 \hat{i}+3 \hat{j}) \cdot(3 \hat{i}-3 \hat{j}-3 \hat{k})\)
= 9 - 9 = 0.
Hence, shortest distance between the given lines is 0.
(v) (c): Since, the point (1, 2, -1) satisfy both the equations of lines, therefore point of intersection of given lines is (1, 2, -1). So, the motorcycles will meet with an accident at the point (1, 2, -1).

(i) (b) : \(\overrightarrow{A B}=(-2 \hat{i}+4 \hat{j}+\hat{k})-(\hat{i}+\hat{j}+\hat{k})=-3 \hat{i}+3 \hat{j}\)
\(\therefore \overrightarrow{A B}=\sqrt{(-3)^{2}+3^{2}}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}\)
Distance between House (A) and ATM (B) is 3\(\sqrt 2\) units.
(ii) (c): \(\overrightarrow{B C}=(-\hat{i}+5 \hat{j}+5 \hat{k})-(-2 \hat{i}+4 \hat{j}+\hat{k})=\hat{i}+\hat{j}+4 \hat{k}\)
\( \therefore |\overrightarrow{B C}| =\sqrt{1^{2}+1^{2}+4^{2}}=\sqrt{1+1+16} \)
\(=\sqrt{18}=3 \sqrt{2}\)
Distance between ATM (B) and School (C) is 3\(\sqrt 2\) units.
(iii) (a): \(\overrightarrow{C D}=(2 \hat{i}+2 \hat{j}+5 \hat{k})-(-\hat{i}+5 \hat{j}+5 \hat{k})=3 \hat{i}-3 \hat{j}\)
\(\therefore |\overrightarrow{C D}|=\sqrt{3^{2}+(-3)^{2}}=\sqrt{9+9}=3 \sqrt{2}\)
Distance between School (C) and Shopping mall (D) is 3\(\sqrt 2\) units.
(iv) (d): Total distance travelled by Ritika
\( =|\overrightarrow{A B}|+|\overrightarrow{B C}|+|\overrightarrow{C D}|=(3 \sqrt{2}+3 \sqrt{2}+3 \sqrt{2}) \text { units } \)
\(=9 \sqrt{2} \text { units }\)
(v) (c): Distance between house and shopping mall is \(|\overrightarrow{A D}|\)
Now, \(\overrightarrow{A D}=\hat{i}+\hat{j}+4 \hat{k}\)
\(\therefore|\overrightarrow{A D}|=\sqrt{1^{2}+1^{2}+4^{2}}=\sqrt{1+1+16}=\sqrt{18}=3 \sqrt{2}\)
Thus, extra distance travelled by Ritika in reaching shopping mall = \((9 \sqrt{2}-3 \sqrt{2})\) units = \(6 \sqrt{2} \) units.