Given, \(\vec{a}+\vec{b}+\vec{c}=0\)
\((\vec{a}+\vec{b}+\vec{c})^2=|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2\)\(+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})\)
\(\begin{array}{ll} \Rightarrow & 0=9+25+16+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}) \end{array}\)
\(\begin{array}{ll} \Rightarrow & -\frac{50}{2}=(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}) \end{array}\)
\(\begin{array}{ll} \therefore & (\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=-25 \end{array}\)

\(\vec{a} \times \vec{b}=\left|\begin{array}{rrr} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 2 & 3 & -5 \end{array}\right|=\hat{i}+11 \hat{j}+7 \hat{k} \)
\(\vec{a} \cdot(\vec{a} \times \vec{b})=1-22+21=0 \)
\(\text { As dot product of } \vec{a} \text { and } \vec{a} \times \vec{b} \text { is zero. Hence, } \vec{a}\text { is perpendicular to } \vec{a} \times \vec{b} \text { . }\)

Let \(\vec{a}, \vec{b} \text { and } \vec{c}\) be the position vectors of points \(P, Q \text { and } R\) . respectively. Then, \(\vec{a}=-\hat{i}+2 \hat{j}-\hat{k}, \vec{b}=3 \hat{i}-\hat{j}+2 \hat{k}\) and \(\vec{c}=2 \hat{i}+3 \hat{j}-\hat{k}\) .
Clearly, the area of \(\Delta P Q R=\frac{1}{2}|\overrightarrow{P Q} \times \overrightarrow{P R}|\) 
Now,\(\overrightarrow{P Q}=\text { Position vector of } Q-\text { Position vector of } P\) 
\(=\vec{b}-\vec{a}=(3 \hat{i}-\hat{j}+2 \hat{k})-(-\hat{i}+2 \hat{j}-\hat{k})\) 
\(=4 \hat{i}-3 \hat{j}+3 \hat{k}\) 
\(\overrightarrow{P R}=\text { Position vector of } R \text { - Position vector of } P\) 
\(=\vec{c}-\vec{a}=(2 \hat{i}+3 \hat{j}-\hat{k})-(-\hat{i}+2 \hat{j}-\hat{k})=3 \hat{i}+\hat{j}\) 
\(\therefore \overrightarrow{P Q} \times \overrightarrow{P R}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 4 & -3 & 3 \\ 3 & 1 & 0 \end{array}\right|=(0-3) \hat{i}-(0-9) \hat{j}+(4+9) \hat{k}\) 
\(=-3 \hat{i}+9 \hat{j}+13 \hat{k}\) 
and \(|\overrightarrow{P Q} \times \overrightarrow{P R}|=\sqrt{(-3)^{2}+(9)^{2}+(13)^{2}}\) 
\(=\sqrt{9+81+169}=\sqrt{259}\) 
So, area of \(\Delta P Q R=\frac{1}{2}|\overrightarrow{P Q} \times \overrightarrow{P R}|=\frac{1}{2} \sqrt{259}\)

Since, \(\vec{\beta}_{1} \text { is parallel to } \vec{\alpha}\) 
Therefore, \(\vec{\beta}_{1}=\lambda \vec{\alpha} \text { for some scalar } \lambda\) 
It is given that, \(\vec{\beta}=\vec{\beta}_{1}+\vec{\beta}_{2}\)
\(\Rightarrow \vec{\beta}_{2}=\vec{\beta}-\vec{\beta}_{1}=\vec{\beta}-\lambda \vec{\alpha}\) 
Also, it is given that \(\vec{\beta}_{2}\) is perpendicular to \(\vec{\alpha}\),therefore
\(\Rightarrow \vec{\beta}_{2} \cdot \vec{\alpha}=0\) 
\(\Rightarrow (\vec{\beta}-\lambda \vec{\alpha}) \cdot \vec{\alpha}=0\) 
\(\Rightarrow \vec{\beta} \cdot \vec{\alpha}-\lambda(\vec{\alpha} \cdot \vec{\alpha})=0\) [using Eq. (ii)]
\(\Rightarrow \lambda=\frac{\vec{\beta} \cdot \vec{\alpha}}{\vec{\alpha} \cdot \vec{\alpha}}\) 
Also,\(\text { given } \vec{\alpha}=3 \hat{i}-\hat{j} \text { and } \vec{\beta}=2 \hat{i}+\hat{j}-3 \hat{k}\) 
Now,\(\overrightarrow{\boldsymbol{\beta}} \cdot \vec{\alpha}=(2 \hat{i}+\hat{j}-3 \hat{k}) \cdot(3 \hat{i}-\hat{j})=6-1+0=5\) 
and \(\vec{\alpha} \cdot \vec{\alpha}=(3 \hat{i}-\hat{j}) \cdot(3 \hat{i}-\hat{j})=9+1=10\) 
\(\therefore \lambda=\frac{\vec{\beta} \cdot \vec{\alpha}}{\vec{\alpha} \cdot \vec{\alpha}}=\frac{5}{10}=\frac{1}{2}\) 
From Equation (i), we get
\(\vec{\beta}_{1}=\lambda \vec{\alpha} \Rightarrow \vec{\beta}_{1}=\frac{1}{2}(3 \hat{i}-\hat{j})=\frac{3}{2} \hat{i}-\frac{1}{2} \hat{j}\) 
and \(\overrightarrow{\beta_{2}}=\vec{\beta}-\vec{\beta}_{1}\) 
\(\Rightarrow \vec{\beta}_{2}=(2 \hat{i}+\hat{j}-3 \hat{k})-\left(\frac{3}{2} \hat{i}-\frac{1}{2} \hat{j}\right)=\frac{1}{2} \hat{i}+\frac{3}{2} \hat{j}-3 \hat{k}\) 
Hence, \(\vec{\beta}_{1}=\frac{3}{2} \hat{i}-\frac{1}{2} \hat{j} \text { and } \vec{\beta}_{2}=\frac{1}{2} \hat{i}+\frac{3}{2} \hat{j}-3 \hat{k}\)

Let \(\overset\rightarrow c=x\overset\wedge i+y\overset\wedge j+z\overset\wedge k\)
Given\(\overrightarrow a=\overset\wedge i+\overset\wedge j+\overset\wedge k\)
\(\overrightarrow b=\overset\wedge j-\overset\wedge k\)
According to the questions,
\(\overrightarrow a.\overrightarrow c=3\)
\(\Rightarrow(\overset\wedge i+\overset\wedge j+\overset\wedge k).(x\overset\wedge i+y\overset\wedge j+z\overset\wedge k)=3\)
\(\Rightarrow x+y+z=3 ....(i)\)
and \(\overrightarrow a\times \overrightarrow c=\overrightarrow b\)
\(\Rightarrow \left| \begin{matrix} \overset { \wedge }{ i } & \overset { \wedge }{ j } & \overset { \wedge }{ k } \\ 1 & 1 & 1 \\ x & y & z \end{matrix} \right| =\overrightarrow b=\overset\wedge j-\overset\wedge k\)
\(\Rightarrow (z-y)\overset\wedge i+(x-z)\overset\wedge j+(y-x)\overset\wedge k=\overset\wedge j-\overset\wedge k\)
On equating the coefficients of like terms, we get
\(z-y=0,\Rightarrow y=z ...(ii) ​​​​​​​\)
\(x-z=1 ...(iii)\)
and \(y-x=-1 ...(iv)\)
Solving eqns.(i), (ii), (iii) and (iv), we get
\(x=\frac{5}{3}\)
and \(y=2/3=z\)
Hence, \(\overrightarrow c=\frac{5}{3}\overset\wedge i+\frac{2}{3}\overset\wedge j+\frac{2}{3}\overset\wedge k\)

Since \(\vec{\beta}_1=\lambda \vec{\alpha}, \lambda\) is a scalar, i.e., \(\vec{\beta}_1=3 \lambda \hat{i}-\lambda \hat{j}\)
\(\vec{\beta}_2=\vec{\beta}-\vec{\beta}_1=(2-3 \lambda) \hat{i}+(1+\lambda) \hat{j}-3 \hat{k}\) 
Now, since \(\vec{\beta}_2\) is to be perpendicular to a \(\vec{\alpha}\) we should have \(\vec{\alpha} \cdot \vec{\beta}_2=0 \text {. i.e., }\)
\( 3(2-3 \lambda)-(1+\lambda) =0 \)
\(\lambda =\frac{1}{2} \)
\(\vec{\beta}_1 =\frac{3}{2} \hat{i}-\frac{1}{2} \hat{j} \text { and } \vec{\beta}_2=\frac{1}{2} \hat{i}+\frac{3}{2} \hat{j}-3 \hat{k} \)

(i) (c) We have, \(\overrightarrow{A B}=\hat{a} \text { and } \overrightarrow{B C}=\hat{b}\)
By triangle law of addition of vectors,
\(\overrightarrow{A C}=\overrightarrow{A B}+\overrightarrow{B C}=\hat{a}+\hat{b}\)
(ii) (b) We know that AD|| BC
\(\therefore \quad \overrightarrow{A D}=2 \overrightarrow{B C}=2 \hat{b}\)
(iii) (c) By triangle law, we have
\(\begin{aligned} \overrightarrow{A C}+\overrightarrow{C D} & =\overrightarrow{A D} \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad \overrightarrow{C D} & =\overrightarrow{A D}-\overrightarrow{A C} \end{aligned}\)
\(\begin{aligned} =2 \hat{b}-(\hat{a}+\hat{b})=\hat{b}-\hat{a} \end{aligned}\)
(iv) (d) \(\begin{aligned} \overrightarrow{E F}=-\overrightarrow{B C}=-\hat{b} \end{aligned}\)
(v) (a) \(\begin{aligned} \overrightarrow{F A}=-\overrightarrow{C D}=-(\hat{b}-\hat{a})=\hat{a}-\hat{b} \end{aligned}\)

(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.

(c) A is correct; R is incorrect

(a) Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of the Assertion (A).

Given, \(\vec{a}+\vec{b}+\vec{c}=0 \Rightarrow \vec{a}+\vec{b}=-\vec{c}\)
\(\Rightarrow \quad(\vec{a}+\vec{b})^2=(-\vec{c})^2\)   [on squaring both sides]
\(\begin{array}{ll} \Rightarrow & (\vec{a}+\vec{b})(\vec{a}+\vec{b})=(-\vec{c}) \cdot(-\vec{c}) \end{array}\)
\(\begin{array}{ll} \Rightarrow \vec{a} \cdot \vec{a}+\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{a}+\vec{b} \cdot \vec{b}=\vec{c} \cdot \vec{c} \end{array}\)
\(\Rightarrow \quad|\vec{a}|^2+2 \vec{a} \cdot \vec{b}+|\vec{b}|^2=|\vec{c}|^2 \quad[\because \vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{a}]\)
\(\Rightarrow|\vec{a}|^2+2|\vec{a}||\vec{b}| \cos \theta+|\vec{b}|^2=|\vec{c}|^2\)    ...(i)
                              \([\because \vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta]\)
Putting the values of \(|\vec{a}|=5,|\vec{b}|=6 \text { and }|\vec{c}|=9\) in Eq. (i), we get
\(\begin{aligned} (5)^2+2 \times 5 \times 6 \times \cos \theta+(6)^2 & =(9)^2 \end{aligned}\)
\(\begin{aligned} \Rightarrow 25+60 \cos \theta+36 & =81 \end{aligned}\)
\(\begin{aligned} \Rightarrow 60 \cos \theta & =81-61=20 \end{aligned}\)
\(\Rightarrow \quad \cos \theta=\frac{20}{60}=\frac{1}{3} \Rightarrow \theta=\cos ^{-1}\left(\frac{1}{3}\right)\)

To find projection of \((\vec{b}+\vec{c}) \text { on } \vec{a}\)
Given, \(\vec{a}=2 \hat{i}-2 \hat{j}+\hat{k}, \vec{b}=\hat{i}+2 \hat{j}-2 \hat{k}\)
and \(\vec{c}=2 \hat{i}-\hat{j}+4 \hat{k}\)
Consider, \(\begin{aligned} (\vec{b}+\vec{c}) & =(\hat{i}+2 \hat{j}-2 \hat{k})+(2 \hat{i}-\hat{j}+4 \hat{k}) \end{aligned}\)
\(\begin{aligned} =3 \hat{i}+\hat{j}+2 \hat{k} \end{aligned}\)
Now, the projection of \(\vec{b}+\vec{c} \text { on } \vec{a}\) is given by
\(\begin{aligned} \frac{(\vec{b}+\vec{c}) \vec{a}}{|\vec{a}|}=\frac{(3 \hat{i}+\hat{j}+2 \hat{k})(2 \hat{i}-2 \hat{j}+\hat{k})}{\sqrt{2^2+(-2)^2+1^2}} \end{aligned}\)
\(\begin{aligned} \frac{6-2+2}{\sqrt{4+4+1}}=\frac{6}{\sqrt{9}}=\frac{6}{3}=2 \end{aligned}\)

We have,
\(\overrightarrow{A B}\) = (position vector of B) - (position vector of A)
\(=(\hat{i}-3 \hat{j}-5 \hat{k})-(2 \hat{i}-\hat{j}+\hat{k})=-\hat{i}-2 \hat{j}-6 \hat{k}\)
\(\begin{aligned} \overrightarrow{B C}=(3 \hat{i}-4 \hat{j}-4 \hat{k})-(\hat{i}-3 \hat{j}-5 \hat{k})=2 \hat{i}-\hat{j}+\hat{k} \end{aligned}\)
and \(\begin{aligned} \overrightarrow{C A}=(2 \hat{i}-\hat{j}+\hat{k})-(3 \hat{i}-4 \hat{j}-4 \hat{k}) \end{aligned}\)
\(=-\hat{i}+3 \hat{j}+5 \hat{k}\)
Here, \(\overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{C A}=0\)
\(\Rightarrow\) A, B and C are the vertices of a triangle.
Now, \(\overrightarrow{B C} \cdot \overrightarrow{C A}=(2 \hat{i}-\hat{j}+\hat{k}) \cdot(-\hat{i}+3 \hat{j}+5 \hat{k})\)
= - 2 - 3 + 5 = 0
\(\Rightarrow \overrightarrow{B C} \perp \overrightarrow{C A} \Rightarrow \angle C=90^{\circ}\)

Now, area of \(\Delta A B C=\frac{1}{2}|\overrightarrow{C A} \times \overrightarrow{B C}|\)
\(=\frac{1}{2}\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ -1 & 3 & 5 \\ 2 & -1 & 1 \end{array}\right|=\frac{1}{2}|(8 \hat{i}-11 \hat{j}-5 \hat{k})|\)
\(=\frac{1}{2} \sqrt{64+121+25}\)
\(=\frac{1}{2} \sqrt{210}\) sq units

According to given condition, \(|\vec{a}|=|\vec{b}|=|\vec{c}|=\lambda\)   (say)    ...(i)
and \(\vec{a} \cdot \vec{b}=0, \vec{b} \cdot \vec{c}=0 \text { and } \vec{c} \cdot \vec{a}=0\)                         ...(ii)
Now, \(|\vec{a}+\vec{b}+\vec{c}|^2=|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2\) \(+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})\)
\(=\lambda^2+\lambda^2+\lambda^2+2(0+0+0)=3 \lambda^2\)
\(\Rightarrow \quad|\vec{a}+\vec{b}+\vec{c}|=\sqrt{3} \lambda\)
[length cannot be negative]
Suppose, \((\vec{a}+\vec{b}+\vec{c})\) is inclined at angles \(\theta_1, \theta_2\) and \(\theta_3\) respectively with vectors \(\vec{a}, \vec{b} \text { and } \vec{c}\), then
\(\begin{aligned} (\vec{a}+\vec{b}+\vec{c}) \cdot \vec{a}=|\vec{a}+\vec{b}+\vec{c}||\vec{a}| \cos \theta_1 \end{aligned}\)
\(\begin{aligned} {[\because \vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta]} \end{aligned}\)
\(\begin{array}{lr} \Rightarrow & |\vec{a}|^2+\vec{a} \cdot \vec{b}+\vec{a} \cdot \vec{c}=\sqrt{3} \lambda \times \lambda \cos \theta_1 \end{array}\)
\(\begin{array}{lr} \Rightarrow & \lambda^2+0+0=\sqrt{3} \lambda^2 \cos \theta_1 \end{array}\)
                                [from Eqs. (i) and (ii)]
\(\begin{array}{ccc} \therefore & \cos \theta_1=\frac{1}{\sqrt{3}} \end{array}\)
\(\begin{array}{ccc} (\vec{a}+\vec{b}+\vec{c}) \cdot \vec{b}=|\vec{a}+\vec{b}+\vec{c} \|| \vec{b} \mid \cos \theta_2 \\ \end{array}\)
\(\begin{array}{ccc} \Rightarrow & \vec{a} \cdot \vec{b}+|\vec{b}|^2+\vec{c} \cdot \vec{b}=\sqrt{3} \lambda \cdot \lambda \cos \theta_2 \end{array}\)
\(\begin{array}{ccc} \Rightarrow & 0+\lambda^2+0=\sqrt{3} \lambda^2 \cos \theta_2 \end{array}\)
                                       [from Eqs. (i) and (ii)]
\(\Rightarrow \quad \cos \theta_2=\frac{1}{\sqrt{3}}\)
Similarly, \((\vec{a}+\vec{b}+\vec{c}) \cdot \vec{c}=|\vec{a}+\vec{b}+\vec{c}||\vec{c}| \cos \theta_3\)
\(\Rightarrow \quad \cos \theta_3=\frac{1}{\sqrt{3}}\)
Thus, \(\cos \theta_1=\cos \theta_2=\cos \theta_3=\frac{1}{\sqrt{3}}\)
Hence, it is proved that \((\vec{a}+\vec{b}+\vec{c})\) is equally inclined with the vectors \(\vec{a}, \vec{b} \text { and } \vec{c}\).
Hence proved.