Vector Algebra

$\begin{aligned} & \because \quad \forall \text { two vectors } \vec{c} \text { & } \vec{d} \\ & |\vec{c} \times \vec{d}|^2=|\vec{c}|^2|\vec{d}|^2-(\vec{c} \cdot \vec{d})^2 \\ & \text { replacing } \vec{c}=\vec{a} ~\& ~\vec{d}=\vec{r} \\ & \Rightarrow|\vec{a} \times \vec{r}|=|\vec{a}|^2|\vec{r}|^2-(\vec{a} \cdot \vec{r})^2 \\ & \Rightarrow|\vec{a} \times \vec{b}|=|\vec{a}|^2|\vec{r}|^2 \quad(\because \vec{a} \times \vec{r}=\vec{a} \times \vec{b} \text { and } \vec{a} \cdot \vec{r}=0) \end{aligned}$

$\begin{aligned} & \Rightarrow 35=14|\vec{r}|^2 \\ & \Rightarrow|\vec{r}|=\sqrt{\frac{35}{14}}=\sqrt{\frac{5}{2}} \neq \sqrt{10} \end{aligned}$

$\therefore$ Statement I is incorrect

Statement II is correct

$\text { (i.e., } \cos 2 A+\cos 2 B+\cos 2 C \geq-\frac{3}{2} \text { ) }$

Proof: $\because(\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}) \geq 0\quad \text{..... (1)}$

and $|\overrightarrow{O A}|^2=|\overrightarrow{O B}|^2=|\overrightarrow{O C}|^2=R^2\quad \text{..... (2)}$

Now, using (1), we get

$|\overrightarrow{O A}|^2+|\overrightarrow{O B}|^2+|\overrightarrow{O C}|^2 +2(\overrightarrow{O A} \cdot \overrightarrow{O B}+\overrightarrow{O B} \cdot \overrightarrow{O C}+\overrightarrow{O C} \cdot \overrightarrow{O A}) \geq 0$

$ \begin{aligned} & \Rightarrow 3 R^2+2 R^2(\cos 2 A+\cos 2 B+\cos 2 C) \geq 0 \\ & \Rightarrow \cos 2 A+\cos 2 B+\cos 2 C \geq-\frac{3}{2} \end{aligned} $

Area of parallelogram whose diagonals are $\vec{a}+\vec{b}$ and $\vec{b}+\vec{c}$ is

$\begin{aligned} & =\frac{1}{2}|(\vec{a}+\vec{b}) \times(\vec{b}+\vec{c})| \\ & =\frac{1}{2}|\vec{a} \times \vec{b}+\vec{a} \times \vec{c}+\vec{b} \times \vec{c}| \\ & =\frac{1}{2}|-2 \beta \hat{i}-2 \hat{j}+(\alpha+\beta) \hat{k}| \end{aligned}$

$=\frac{1}{2} \sqrt{4 \beta^2+4+(\alpha+\beta)^2}$

Which is given $\frac{\sqrt{21}}{2}$

$\begin{array}{ll} \therefore & 4 \beta^2+4+(\alpha+\beta)^2=21 \\ \Rightarrow & (\alpha+\beta)^2+4 \beta^2=17 \\ \Rightarrow & \alpha^2+5 \beta^2+2 \alpha \beta=17 \\ \Rightarrow & \alpha^2+5 \beta^2=29 \\ \therefore & (\alpha, \beta) \in\{(3,2),(-3,-2),(-3,2),(3,-2)\} \\ \because & \alpha \beta=-6 \\ \therefore & (\alpha, \beta) \in\{(-3,2),(3,-2)\} \\ \therefore & \alpha_1^2+\beta_1^2-\alpha_2 \beta_2 \\ = & 9+4-(-6)=19 \end{array}$

To solve this, let's start with the given vector equation:

$\vec{c} = \vec{a} - \vec{b}$

Given vectors are:

$\overrightarrow{\mathrm{a}} = \alpha \hat{i} + 4 \hat{j} + 2 \hat{k}$

$\overrightarrow{\mathrm{b}} = 5 \hat{i} + 3 \hat{j} + 4 \hat{k}$

Then, the vector $\overrightarrow{\mathrm{c}}$ is:

$\overrightarrow{\mathrm{c}} = (\alpha \hat{i} + 4 \hat{j} + 2 \hat{k}) - (5 \hat{i} + 3 \hat{j} + 4 \hat{k})$

$\overrightarrow{\mathrm{c}} = (\alpha - 5) \hat{i} + (4 - 3) \hat{j} + (2 - 4) \hat{k}$

$\overrightarrow{\mathrm{c}} = (\alpha - 5) \hat{i} + 1 \hat{j} - 2 \hat{k}$

The area of the triangle formed by vectors $\vec{a}$ and $\vec{b}$ is given by the magnitude of the cross product of $\vec{a}$ and $\vec{b}$, divided by 2:

$\text{Area} = \frac{1}{2} |\vec{a} \times \vec{b}| = 5 \sqrt{6}$

This implies:

$|\vec{a} \times \vec{b}| = 10 \sqrt{6}$

Let's find $\vec{a} \times \vec{b}$:

$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \alpha & 4 & 2 \\ 5 & 3 & 4 \end{vmatrix} = \hat{i}(4 \cdot 4 - 2 \cdot 3) - \hat{j}(\alpha \cdot 4 - 2 \cdot 5) + \hat{k}(\alpha \cdot 3 - 4 \cdot 5)$

$\vec{a} \times \vec{b} = \hat{i}(16 - 6) - \hat{j}(4\alpha - 10) + \hat{k}(3\alpha - 20)$

$\vec{a} \times \vec{b} = \hat{i}(10) - \hat{j}(4\alpha - 10) + \hat{k}(3\alpha - 20)$

Magnitude of $\vec{a} \times \vec{b}$:

$|\vec{a} \times \vec{b}| = \sqrt{10^2 + (4\alpha - 10)^2 + (3\alpha - 20)^2}$

We know:

$\sqrt{10^2 + (4\alpha - 10)^2 + (3\alpha - 20)^2} = 10 \sqrt{6}$

Squaring both sides:

$100 + (4\alpha - 10)^2 + (3\alpha - 20)^2 = 600$

$100 + 16\alpha^2 - 80\alpha + 100 + 9\alpha^2 - 120\alpha + 400 = 600$

$25\alpha^2 - 200\alpha + 600 = 600$

$25\alpha^2 - 200\alpha = 0$

$\alpha^2 - 8\alpha = 0$

$\alpha(\alpha - 8) = 0$

Since $\alpha$ is a positive real number:

$\alpha = 8$

Then, the vector $\vec{c}$ is:

$\vec{c} = (8 - 5)\hat{i} + 1\hat{j} - 2\hat{k}$

$\vec{c} = 3\hat{i} + 1\hat{j} - 2\hat{k}$

Magnitude squared of $\vec{c}$ is:

$|\vec{c}|^2 = 3^2 + 1^2 + (-2)^2$

$|\vec{c}|^2 = 9 + 1 + 4$

$|\vec{c}|^2 = 14$

Therefore, the correct option is:

Option A: 14

$\begin{aligned} & 6|\vec{a} \times \vec{b}|=15 \\ & \Rightarrow|\vec{a} \times \vec{b}|=\frac{5}{2} \end{aligned}$

Area of quadrilateral $O A B C$

$=$ area of $\triangle O A C+$ area of $\triangle A B C$

$\begin{aligned} & =\frac{15}{2}+\frac{1}{2}|(\overrightarrow{A B} \times \overrightarrow{B C})| \\ & =\frac{15}{2}+\frac{1}{2}|(4 \vec{a}+5 \vec{b}) \times(6 \vec{a}+2 \vec{b})| \\ & =\frac{15}{2}+\frac{1}{2}|(22 \vec{a} \times \vec{b})| \\ & =\frac{15}{2}+11 \times \frac{5}{2} \\ & =\frac{70}{2}=35 \end{aligned}$

$\begin{aligned} & \left.\begin{array}{l} \vec{a}=4 \hat{i}-\hat{j}+\hat{k} \\ \vec{b}=11 \hat{i}-\hat{j}+\hat{k} \end{array}\right] \\ & \vec{a} \cdot \vec{b}=44+1+1=46 \\ & |\vec{a}|^2=18,\left|\vec{b}^2\right|=123 \\ & (\vec{a}+\vec{b}) \times \vec{c}=\vec{c} \times(-2 \vec{a}+3 \vec{b}) \\ & (2 \vec{a}+3 \vec{b}) \cdot \vec{c}=1670 \\ & (\vec{a}+\vec{b}) \times c=(2 \vec{a}-3 \vec{b}) \times \vec{c} \\ & (-\vec{a}+4 \vec{b}) \times \vec{c}=0 \\ & \vec{c}=\lambda(4 \vec{b}-\vec{a}) \\ & (2 \vec{a}+3 \vec{b}) \cdot \lambda(4 \vec{b}-\vec{a})=1670 \\ & \lambda\left(5 \vec{a} \cdot \vec{b}-2|\vec{a}|^2+12|\vec{b}|^2\right)=1670 \\ & \lambda=\frac{1670}{5 \times 46-2 \times 18+12 \times 123} \\ & \lambda=1 \\ & \vec{c}=4 \vec{b}-\vec{a} \\ & =4(11 \hat{i}-\hat{j}+\hat{k})-(4 \hat{i}-\hat{j}+\hat{k}) \\ & =40 \hat{i}-3 \hat{j}+3 \hat{k} \\ & \left|\vec{c}^2\right|=1600+9+9 \\ & =1618 \\ \end{aligned}$

$\begin{aligned} & \vec{a}=\hat{i}+2 \hat{j}+3 \hat{k} \\ & \vec{b}=2 \hat{i}+3 \hat{j}-5 \hat{k} \\ & \vec{c}=3 \hat{i}-\hat{j}+\lambda \hat{k} \\ & \vec{b}+\vec{c}=5 \hat{i}+2 \hat{j}+(\lambda-5) \hat{k} \end{aligned}$

$\vec{r}$ is a unit vector along $\vec{b}+\vec{c}$

$\therefore \quad \vec{r}=\frac{5 \hat{i}+2 \hat{j}+(\lambda-5) \hat{k}}{\sqrt{25+4+(\lambda-5)^2}}$

Now, $\vec{r} \cdot \vec{a}=3$

$\frac{1}{\sqrt{29+(\lambda-5)^2}}[5+4+3(\lambda-5)]=3$

Squaring both sides

$\begin{aligned} & \Rightarrow \quad \frac{1}{29+(\lambda-5)^2}\left[9+3(\lambda-5)^2\right]=9 \\ & \Rightarrow \quad[3+(\lambda-5)]^2=29+(\lambda-5)^2 \\ & \Rightarrow \quad 9+(\lambda-5)^2+6(\lambda-5)=29+(\lambda-5)^2 \\ & \Rightarrow \quad 9+6(\lambda-5)=29 \\ & \Rightarrow \quad \lambda=\frac{20}{6}+5=\frac{25}{3} \\ & \therefore \quad 3 \lambda=3 \times \frac{25}{3}=25 \end{aligned}$

Given $\vec{a}=\alpha t \hat{i}+6 \hat{j}-3 \hat{k}$

and $\vec{b}=t \hat{i}-2 \hat{j}-2 \alpha t \hat{k}$

angle between $\vec{a}$ and $\vec{b}$ is given by

$\cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$

We have, $\cos \theta < 0(\because$ angle between $\vec{a}$ and $\vec{b}$ is obtuse)

$\begin{aligned} & \Rightarrow \quad \vec{a} \cdot \vec{b}<0 \\ & \Rightarrow \alpha t^2-12+6 \alpha t<0 \forall t \in \mathbb{R} \end{aligned}$

If $\alpha=0$, then $-12<0$ (condition holds)

If $\alpha \neq 0 \Rightarrow \alpha<0\quad \text{.... (i)}$

And maximum value of $\alpha t^2+6 \alpha t-12<0$

$\begin{aligned} & \Rightarrow \frac{-D}{4 a}<0 \text { (where } D \text { is discriminant and } a=\alpha) \\ & \Rightarrow \frac{36 \alpha^2+48 \alpha}{4 \alpha}>0 \\ & \Rightarrow \alpha>\frac{-4}{3} \\ & \therefore \alpha \in\left(\frac{-4}{3}, 0\right] \end{aligned}$

$\begin{aligned} & \vec{a}=2 \hat{i}+\hat{j}-\hat{k} \\ & \vec{b}=((\vec{a} \times(\hat{i}+\hat{j})) \times \hat{i}) \times \hat{i} \\ & \vec{a} \times(\hat{i}+\hat{j})=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ 1 & 1 & 0 \end{array}\right| \\ &=\hat{i}(1)-\hat{j}(1)+\hat{k}(2-1) \\ &=\hat{i}-\hat{j}+\hat{k} \end{aligned}$

$\begin{aligned} & (\vec{a} \times(\hat{i}+\hat{j})) \times \hat{i}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 1 & 0 & 0 \end{array}\right| \\ & \quad=\hat{i}(0)-\hat{j}(-1)+\hat{k}(1) \\ & \quad=\hat{j}+\hat{k} \end{aligned}$

$\left( {(\overrightarrow a \times (\widehat i + \widehat j) \times \widehat i} \right) \times \widehat i = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 0 & 1 & 1 \cr 1 & 0 & 0 \cr } } \right|$

$\begin{aligned} & =\hat{i}(0)-\hat{j}(-1)+\hat{k}(-1) \\ \vec{b}= & \hat{j}-\hat{k} \end{aligned}$

Projection of $\vec{a}$ on $\vec{b}=\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$

$=\frac{2}{\sqrt{2}}=\sqrt{2}$

Square of projection $=2$

$\begin{aligned} & |(\vec{a} \times \vec{b}) \times \vec{c}|=|\vec{a} \times \vec{b}||\vec{c}| \sin 60^{\circ} \\ & \left|\begin{array}{ccc} i & j & k \\ 6 & 1 & -1 \\ 1 & 1 & 0 \end{array}\right|=i(1)-j(1)+k(5) \\ & =i-j+5 k \\ & |\vec{a} \times \vec{b}|=\sqrt{1+1+25}=\sqrt{27} \\ & |\vec{c}-\vec{a}|=2 \sqrt{2} \\ & c^2+a^2-2 a c=8 \\ & c^2-12 c+30=0 \\ & c=\frac{12+\sqrt{24}}{2}=6+\sqrt{6} \\ & \Rightarrow|(\vec{a} \times \vec{b}) \times \vec{c}|=\sqrt{27} \times(6+\sqrt{6}) \times \frac{\sqrt{3}}{2} \\ & =\frac{a}{2}(6+\sqrt{6}) \\ \end{aligned}$

$\begin{gathered} (\vec{c}+\hat{i}) \times(\vec{a}+\vec{b}+\hat{i}+\vec{a})=0 \\ \Rightarrow \quad \vec{c}+\hat{i}=\lambda(\vec{a}+\vec{b}+\hat{i}+\vec{a}) \\ =\lambda(2 \vec{a}+\vec{b}+\hat{i}) \\ \quad=\lambda(7 \hat{i}+8 \hat{j}) \\ \Rightarrow \quad \vec{c}=(7 \lambda-1) \hat{i}+8 \lambda \hat{j} \\ \quad \vec{c} \cdot \vec{a}=-29 \\ \Rightarrow \quad 14 \lambda-2+40 \lambda=-29 \\ \Rightarrow \quad 54 \lambda=-27 \\ \Rightarrow \quad \lambda=-\frac{1}{2} \end{gathered}$

$\begin{aligned} & \therefore \quad \vec{c}=\left(\frac{-7}{2}-1\right) \hat{i}-4 \hat{j}=\frac{-9}{2} \hat{i}-4 \hat{j} \\ & \vec{c} \cdot(-2 \hat{i}+\hat{j}+\hat{k})=9-4=5 \end{aligned}$

$\begin{aligned} & \vec{a}=\vec{b} \times \vec{c} \\ & |\vec{a}|=2,|\vec{b}|=3 \end{aligned}$

$\vec{a} \cdot \vec{b}=0$ and $\vec{a} \cdot \vec{c}=0$

$\begin{aligned} & |\vec{c}-\vec{a}|^2=|\vec{c}|^2+|\vec{a}|^2-2 \vec{c} \cdot \vec{a} \\ & =4+|\vec{c}|^2 \end{aligned}$

$\begin{aligned} & |\vec{a}|=|\vec{b} \times \vec{c}|=|\vec{b}| \sin \alpha|\vec{c}| \\ & \Rightarrow \sin \alpha|\vec{c}|=\frac{2}{3} \\ & \Rightarrow \sin ^2 \alpha=\frac{4}{9|\vec{c}|^2} \\ & \Rightarrow|\vec{c}|^2=\frac{4}{9 \sin ^2 \alpha} \\ & \Rightarrow|\vec{c}-\vec{a}|^2=4+\frac{4}{9 \sin ^2 \alpha} \end{aligned}$

For $|\vec{c}-\vec{a}|^2$ to be minimum for $\alpha \in\left[0, \frac{\pi}{3}\right]$

$\begin{gathered} \sin \alpha=\frac{\sqrt{3}}{2} \\ 27|\vec{c}-\vec{a}|^2=27\left[4+\frac{4.4}{9.3}\right] \\ =27\left[\frac{124}{27}\right]=124 \end{gathered}$

Area of quadrilateral $A B C D=$ area of $\triangle A B C+$ area of $\triangle A D C$

In $\triangle A B C$

$\begin{aligned} & \overrightarrow{A B}=4,8,-8 \\ & \overrightarrow{B C}=-2,-3,-4 \\ & \overrightarrow{A B} \times \overrightarrow{B C}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 4 & 8 & -8 \\ -2 & -3 & -4 \end{array}\right| \\ & =-56 \hat{i}+32 \hat{j}+4 \hat{k} \end{aligned}$

Area of $\triangle A B C=\frac{1}{2}|\overrightarrow{A B} \times \overrightarrow{B C}|$

$\begin{aligned} & =\frac{1}{2} \sqrt{56^2+32^2+4^2} \\ & =\frac{1}{2} \sqrt{4176}=\frac{12 \sqrt{29}}{2}=6 \sqrt{29} \end{aligned}$

In $\triangle A D C=\overrightarrow{A D}=-2,-3,-4$

$\overrightarrow{D C}=4,8,-8$

$\overrightarrow{A D} \times \overrightarrow{D C}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ -2 & -3 & -4 \\ 4 & 8 & -8\end{array}\right|$

$=56 \hat{i}-32 \hat{j}-4 k$

$\text { Area of } \frac{1}{2}|\overrightarrow{A D} \times \overrightarrow{D C}|$

$\begin{aligned} & =\frac{1}{2} \sqrt{4176} \\ & =6 \sqrt{29} \end{aligned}$

Area of $A B C D=6 \sqrt{29}+6 \sqrt{29}$

$=12 \sqrt{29}$

We are given the vectors

$ \vec{a} = \hat{i} + \lambda\,\hat{j} - 3\,\hat{k} \quad \text{and} \quad \vec{b} = 3\,\hat{i} - \hat{j} + 2\,\hat{k}, $

with $\lambda > 0$, and we are told that the vectors $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ are perpendicular. Our goal is to find $(14 \cos\theta)^2$, where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$.

Step 1. Use the Perpendicularity Condition

Two vectors are perpendicular if their dot product is zero. So, we require:

$ (\vec{a}+\vec{b})\cdot (\vec{a}-\vec{b})=0. $

First, compute $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$:

$ \vec{a}+\vec{b} = (\hat{i}+3\,\hat{i})+(\lambda\,\hat{j}-\hat{j})+(-3\,\hat{k}+2\,\hat{k}) = 4\,\hat{i}+(\lambda-1)\,\hat{j}-\hat{k}. $

$ \vec{a}-\vec{b} = (\hat{i}-3\,\hat{i})+(\lambda\,\hat{j}+ \hat{j})+(-3\,\hat{k}-2\,\hat{k}) = -2\,\hat{i}+(\lambda+1)\,\hat{j}-5\,\hat{k}. $

Now, take their dot product:

$ (4\,\hat{i}+(\lambda-1)\,\hat{j}-\hat{k}) \cdot (-2\,\hat{i}+(\lambda+1)\,\hat{j}-5\,\hat{k}) $

$ = 4(-2) + (\lambda-1)(\lambda+1) + (-1)(-5) = -8 + (\lambda^2 - 1) + 5. $

$ = \lambda^2 - 4. $

Setting the dot product to zero:

$ \lambda^2 - 4 = 0 \quad \Longrightarrow \quad \lambda^2 = 4. $

Since $\lambda > 0$, we have:

$ \lambda = 2. $

Step 2. Find the Angle Between $\vec{a}$ and $\vec{b}$

Now that we know $\lambda = 2$, the vectors become:

$ \vec{a} = \hat{i} + 2\,\hat{j} - 3\,\hat{k}, \quad \vec{b} = 3\,\hat{i} - \hat{j} + 2\,\hat{k}. $

The cosine of the angle $\theta$ between $\vec{a}$ and $\vec{b}$ is given by:

$ \cos\theta = \frac{\vec{a}\cdot\vec{b}}{\|\vec{a}\| \|\vec{b}\|}. $

Calculate $\vec{a} \cdot \vec{b}$:

$ \vec{a}\cdot\vec{b} = (1)(3) + (2)(-1) + (-3)(2) = 3 - 2 - 6 = -5. $

Calculate the magnitudes:

$ \|\vec{a}\| = \sqrt{1^2 + 2^2 + (-3)^2} = \sqrt{1 + 4 + 9} = \sqrt{14}. $

$ \|\vec{b}\| = \sqrt{3^2 + (-1)^2 + 2^2} = \sqrt{9 + 1 + 4} = \sqrt{14}. $

Thus,

$ \cos\theta = \frac{-5}{\sqrt{14}\cdot\sqrt{14}} = \frac{-5}{14}. $

Step 3. Compute $(14\cos\theta)^2$

$ (14\cos\theta)^2 = \left(14\cdot\frac{-5}{14}\right)^2 = (-5)^2 = 25. $

Final Answer

The value of $(14\cos\theta)^2$ is:

$ \boxed{25}. $

$\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=2 \hat{i}+4 \hat{j}-5 \hat{k}, \vec{c}=x \hat{i}+2 \hat{j}+3 \hat{k}, x \in R$

$\text { also, } \vec{b}+\vec{c}=(x+2) \hat{i}+6 \hat{j}-2 \hat{k}$

$\vec{d} \text { is the unit vector in the direction of } \vec{b}+\vec{c}$

$\begin{aligned} & |\vec{b}+\vec{c}|=\sqrt{(x+2)^2+6^2+2^2} \\ & =\sqrt{40+(x+2)^2} \\ & \vec{d}=\frac{x+2}{\sqrt{40+(x+2)^2}} \hat{i}+\frac{6}{\sqrt{40+(x+2)^2}} \hat{j} -\frac{2}{\sqrt{40+(x+2)^2}} \hat{k} \\ \end{aligned}$

$\begin{aligned} & \vec{a} \cdot \vec{d}=1 \\ & \frac{x+2+6-2}{\sqrt{40+(x+2)^2}}=1 \\ & x+6=\sqrt{40+(x+2)^2} \end{aligned}$

$\begin{aligned} &(x+6)^2=40+(x+2)^2 \\ & x^2+36+ 12 x=40+x^2+4+4 x \\ & 8 x=8 \\ & \Rightarrow x=1 \\ &(\vec{a} \times \vec{b}) \cdot \vec{c}=[\vec{a} \vec{b} \vec{c}] \\ &=\left[\begin{array}{ccc} 1 & 1 & 1 \\ 2 & 4 & -5 \\ 1 & 2 & 3 \end{array}\right] \\ &=1(12+10)-1(6+5)+1(4-4) \\ &=22-11=11 \end{aligned}$

$\begin{aligned} & \text { Let } \vec{C}=a \hat{i}+b \hat{j}+c \hat{k} \\ & (a \hat{i}+b \hat{j}+c \hat{k}) \cdot(2 \hat{i}+2 \hat{j}-\hat{k})=1 \times 3 \times \frac{1}{2} \\ & 2 a+2 b-c=\frac{3}{2} \qquad \text{... (1)}\\ & (a \hat{i}+b \hat{j}+c \hat{k}) \cdot(\hat{i}-\hat{k})=1 \times \sqrt{2} \times \frac{1}{\sqrt{2}} \\ & a-c=1 \quad \text{... (2)}\\ & a^2+b^2+c^2=1 \quad \text{... (3)}\\ \end{aligned}$

Solving (1), (2) and (3)

$\begin{aligned} & a+2 b=\frac{1}{2} \\ & a^2+b^2+(a-1)^2=1 \\ & 2 a^2-2 a+b^2=0 \\ & 2 a^2-2 a+\left(\frac{2 a-1}{4}\right)^2=0 \\ & 32 a^2-32 a+4 a^2-4 a+1=0 \\ & 36 a^2-36 a+1=0 \\ & a=\frac{36 \pm \sqrt{(36)^2-4(36)}}{2 \times 36} \\ & =\frac{1}{2} \pm \frac{\sqrt{2}}{3} \\ & b=\frac{1-2 a}{4} \Rightarrow b=\frac{1 \pm \frac{2 \sqrt{2}}{3}-1}{4} \\ & =\mp \frac{1}{3 \sqrt{2}} \\ & C=-\frac{1}{2} \pm \frac{\sqrt{2}}{3} \end{aligned}$

$\begin{aligned} & C+\left(\frac{-1}{2} \hat{i}+\frac{1}{3 \sqrt{2}} \hat{j}-\frac{\sqrt{2}}{3} \hat{k}\right) \\ & =\frac{\sqrt{2}}{3} \hat{i}-\frac{1}{2} \hat{k} \end{aligned}$

$\begin{aligned} & \vec{a}=-5 \cdot \hat{i}+\hat{j}-3 \hat{k}, \vec{b}=\hat{i}+2 \hat{j}-4 \hat{k} \\\\ & \vec{c}=(((\vec{a} \times \vec{b}) \times \hat{i}) \times \hat{i}) \times \hat{i} \\\\ & =(((\vec{a} \cdot \hat{i}) \vec{b}-(\vec{b} \cdot \hat{i}) \vec{a}) \times \hat{i}) \times \hat{i} \\\\ & =((-5 \vec{b}-\vec{a}) \times \hat{i}) \times \hat{i}\end{aligned}$

$\begin{aligned} & =((-11 \hat{j}+23 \hat{k}) \times \hat{i}) \times \hat{i} \\\\ & = (11 \hat{k}+23 \hat{j}) \times \hat{i} \\\\ & = (11 \hat{j}-23 \hat{k})\end{aligned}$

$\begin{aligned} & \vec{c} \cdot(-\hat{i}+\hat{j}+\hat{k})=0+11-23 \\\\ & =-12\end{aligned}$

$\begin{aligned} & \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{b}}=\overrightarrow{0} \\ & (\overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{c}}) \times \overrightarrow{\mathrm{b}}=\overrightarrow{0} \\ & \overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{c}}=\lambda \overrightarrow{\mathrm{b}} \Rightarrow \overrightarrow{\mathrm{p}}=\overrightarrow{\mathrm{c}}+\lambda \overrightarrow{\mathrm{b}} \end{aligned}$

Now, $\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{a}}=0$ (given)

$\begin{aligned} & \text { So, } \overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=0 \\ & (3-3-8)+\lambda(12+1-14)=0 \\ & \lambda=-8 \\ & \overrightarrow{\mathrm{p}}=\overrightarrow{\mathrm{c}}-8 \overrightarrow{\mathrm{b}} \\ & \overrightarrow{\mathrm{p}}=-31 \hat{\mathrm{i}}-11 \hat{\mathrm{j}}-52 \hat{\mathrm{k}} \\ & \text { So, } \overrightarrow{\mathrm{p}} \cdot(\hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}}) \\ & =-31+11+52 \\ & =32 \end{aligned}$

A vector in the direction of the required line can be obtained by cross product of

$\begin{aligned} & \left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 5 \\ -1 & 3 & 2 \end{array}\right| \\\\ & =-9 \hat{i}-9 \hat{j}+9 \hat{k} \end{aligned}$

Required line

$\begin{aligned} & \overrightarrow{\mathrm{r}}=(5 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\lambda^{\prime}(-9 \hat{\mathrm{i}}-9 \hat{\mathrm{j}}+9 \hat{\mathrm{k}}) \\\\ & \overrightarrow{\mathrm{r}}=(5 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\lambda(\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}) \end{aligned}$

Now distance of $(0,2,-2)$

$\begin{aligned} & \text { P.V. of } \mathrm{P} \equiv(5+\lambda) \hat{i}+(\lambda-4) \hat{j}+(3-\lambda) \hat{k} \\\\ & \overrightarrow{\mathrm{AP}}=(5+\lambda) \hat{i}+(\lambda-6) \hat{j}+(5-\lambda) \hat{k} \\\\ & \overrightarrow{\mathrm{AP}} \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})=0 \\\\ & 5+\lambda+\lambda-6-5+\lambda=0 \\\\ & \lambda=2 \\\\ & |\overrightarrow{\mathrm{AP}}|=\sqrt{49+16+9} \\\\ & |\overrightarrow{\mathrm{AP}}|=\sqrt{74} \end{aligned}$

$\begin{aligned} & |\overrightarrow{\mathrm{b}}|^2=6 ;|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}| \cos \theta=3 \sqrt{2} \\ & |\overrightarrow{\mathrm{a}}|^2|\overrightarrow{\mathrm{b}}|^2 \cos ^2 \theta=18 \\ & |\overrightarrow{\mathrm{a}}|^2=6 \end{aligned}$

Also $1+\alpha^2+\beta^2=6$

$\alpha^2+\beta^2=5$

to find

$\begin{aligned} & \left(\alpha^2+\beta^2\right)|\vec{a}|^2|\vec{b}|^2 \sin ^2 \theta \\ & =(5)(6)(6)\left(\frac{1}{2}\right) \\ & =90 \end{aligned}$

To find the value of $|(\vec{b} \times \vec{a})-\vec{b}|^2$, we can use properties of vector operations and magnitudes. Given $|\vec{b}| = 1$ and $|\vec{b} \times \vec{a}| = 2$, let's break down the calculation step by step:

Firstly, we observe that the cross product of two vectors $\vec{b} \times \vec{a}$ is orthogonal (perpendicular) to both $\vec{b}$ and $\vec{a}$. This means that when we take the dot product of $\vec{b} \times \vec{a}$ with either $\vec{b}$ or $\vec{a}$, the result will be zero due to the orthogonal property. Specifically,

$(\vec{b} \times \vec{a}) \cdot \vec{b}= 0 $(because $\vec{b} \times \vec{a}$ is perpendicular to both $\vec{b}$ and $\vec{a}$)

Next, to find the magnitude squared of the vector $(\vec{b} \times \vec{a}) - \vec{b}$, we apply the formula for the magnitude squared of a vector subtraction, which can be expressed as:

$|(\vec{b} \times \vec{a}) - \vec{b}|^2 = |\vec{b} \times \vec{a}|^2 + |-\vec{b}|^2 + 2(\vec{b} \times \vec{a}) \cdot (-\vec{b})$

Since $(\vec{b} \times \vec{a}) \cdot \vec{b} = 0$ and $|-\vec{b}| = |\vec{b}|$, the equation simplifies to:

$= |\vec{b} \times \vec{a}|^2 + |\vec{b}|^2$

Given that $|\vec{b} \times \vec{a}| = 2$ and $|\vec{b}| = 1$, substituting these values gives:

$= 2^2 + 1^2 = 4 + 1 = 5$

Therefore, the value of $|(\vec{b} \times \vec{a})-\vec{b}|^2$ is 5.

Given $|\vec{a}|=1,|\vec{b}|=4, \vec{a} \cdot \vec{b}=2$

$\vec{\mathrm{c}}=2(\vec{\mathrm{a}} \times \vec{\mathrm{b}})-3 \vec{\mathrm{b}}$

Dot product with $\overrightarrow{\mathrm{a}}$ on both sides

$\overrightarrow{\mathrm{c}} . \overrightarrow{\mathrm{a}}=-6$ ..... (1)

Dot product with $\vec{b}$ on both sides

$\begin{aligned} & \overrightarrow{\mathrm{b}} \overrightarrow{\mathrm{c}}=-48 \quad \text{... (2)}\\ & \overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{c}}=4|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|^2+9|\overrightarrow{\mathrm{b}}|^2 \\ & |\overrightarrow{\mathrm{c}}|^2=4\left[|\mathrm{a}|^2|\mathrm{~b}|^2-(\mathrm{a} \cdot \overrightarrow{\mathrm{b}})^2\right]+9|\overrightarrow{\mathrm{b}}|^2 \\ & |\overrightarrow{\mathrm{c}}|^2=4\left[(1)(4)^2-(4)\right]+9(16) \\ & |\overrightarrow{\mathrm{c}}|^2=4[12]+144 \\ & |\overrightarrow{\mathrm{c}}|^2=48+144 \\ & |\overrightarrow{\mathrm{c}}|^2=192 \\ & \therefore \cos \theta=\frac{\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}}{|\overrightarrow{\mathrm{b}}| \overrightarrow{\mathrm{c}} \mid} \\ & \therefore \cos \theta=\frac{-48}{\sqrt{192} \cdot 4} \\ & \therefore \cos \theta=\frac{-48}{8 \sqrt{3} .4} \\ & \therefore \cos \theta=\frac{-3}{2 \sqrt{3}} \\ & \therefore \cos \theta=\frac{-\sqrt{3}}{2} \Rightarrow \theta=\cos ^{-1}\left(\frac{-\sqrt{3}}{2}\right) \end{aligned}$

Unit vector $\hat{\mathrm{u}}=\mathrm{x} \hat{\mathrm{i}}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}}$

$\begin{aligned} & \overrightarrow{\mathrm{p}}_1=\frac{1}{\sqrt{2}} \hat{\mathrm{i}}+\frac{1}{\sqrt{2}} \hat{\mathrm{k}}, \overrightarrow{\mathrm{p}}_2=\frac{1}{\sqrt{2}} \hat{\mathrm{j}}+\frac{1}{\sqrt{2}} \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{p}}_3=\frac{1}{\sqrt{2}} \hat{\mathrm{i}}+\frac{1}{\sqrt{2}} \hat{\mathrm{j}} \end{aligned}$

Now angle between $\hat{\mathrm{u}}$ and $\overrightarrow{\mathrm{p}}_1=\frac{\pi}{2}$

$\hat{\mathrm{u}} \cdot \overrightarrow{\mathrm{p}}_1=0 \Rightarrow \frac{\mathrm{x}}{\sqrt{2}}+\frac{\mathrm{z}}{\sqrt{2}}=0$

$\Rightarrow \mathrm{x}+\mathrm{z}=0$ ...... (i)

Angle between $\hat{\mathrm{u}}$ and $\overrightarrow{\mathrm{p}}_2=\frac{\pi}{3}$

$\hat{\mathrm{u}} \cdot \overrightarrow{\mathrm{p}}_2=|\hat{\mathrm{u}}| \cdot\left|\overrightarrow{\mathrm{p}}_2\right| \cos \frac{\pi}{3}$

$\Rightarrow \frac{y}{\sqrt{2}}+\frac{z}{\sqrt{2}}=\frac{1}{2} \Rightarrow y+z=\frac{1}{\sqrt{2}}$ ...... (ii)

Angle between $\hat{\mathrm{u}}$ and $\overrightarrow{\mathrm{p}}_3=\frac{2 \pi}{3}$

$\hat{\mathrm{u}} \cdot \overrightarrow{\mathrm{p}}_3=|\hat{\mathrm{u}}| \cdot\left|\overrightarrow{\mathrm{p}}_3\right| \cos \frac{2 \pi}{3}$

$\Rightarrow \frac{x}{\sqrt{2}}+\frac{4}{\sqrt{2}}=\frac{-1}{2} \Rightarrow x+y=\frac{-1}{\sqrt{2}}$ ..... (iii)

from equation (i), (ii) and (iii) we get

$x=\frac{-1}{\sqrt{2}} \quad y=0 \quad z=\frac{1}{\sqrt{2}}$

$\begin{aligned} & \text { Thus } \hat{\mathrm{u}}-\overrightarrow{\mathrm{v}}=\frac{-1}{\sqrt{2}} \hat{\mathrm{i}}+\frac{1}{\sqrt{2}} \hat{\mathrm{k}}-\frac{1}{\sqrt{2}} \hat{\mathrm{i}}-\frac{1}{\sqrt{2}} \hat{\mathrm{j}}-\frac{1}{\sqrt{2}} \hat{\mathrm{k}} \\ & \hat{\mathrm{u}}-\overrightarrow{\mathrm{v}}=\frac{-2}{\sqrt{2}} \hat{\mathrm{i}}-\frac{1}{\sqrt{2}} \hat{\mathrm{j}} \\ & \therefore|\hat{\mathrm{u}}-\overrightarrow{\mathrm{v}}|^2=\left(\sqrt{\frac{4}{2}+\frac{1}{2}}\right)^2=\frac{5}{2} \end{aligned}$

Area of parallelogram, $S=|\vec{a} \times \vec{b}|$

Area of quadrilateral $=\operatorname{Area}(\triangle \mathrm{OAB})+\operatorname{Area}(\triangle \mathrm{OBC})$

$\begin{aligned} & =\frac{1}{2}\{|\vec{a} \times(12 \vec{a}+4 \vec{b})|+|\vec{b} \times(12 \vec{a}+4 \vec{b})|\} \\ & =8|(\vec{a} \times \vec{b})| \end{aligned}$

$\text { Ratio }=\frac{8|(\vec{a} \times \vec{b})|}{|(\vec{a} \times \vec{b})|}=8$

$\begin{aligned} & \vec{a}+5 \vec{b}=\lambda \vec{c} \\ & \vec{b}+6 \vec{c}=\mu \vec{a} \end{aligned}$

Eliminating $\vec{a}$

$\begin{aligned} & \lambda \overrightarrow{\mathrm{c}}-5 \overrightarrow{\mathrm{b}}=\frac{6}{\mu} \overrightarrow{\mathrm{c}}+\frac{1}{\mu} \overrightarrow{\mathrm{b}} \\ & \therefore \mu=\frac{-1}{5}, \lambda=-30 \\ & \alpha=5, \beta=30 \end{aligned}$

$\triangle \mathrm{ABC}$ is equilateral

Orthocentre and centroid will be same

$\mathrm{G}\left(\frac{5}{3}, \frac{5}{3}, \frac{5}{3}\right)$

Mid-point of $\mathrm{AB}$ is $\mathrm{D}\left(\frac{3}{2}, 2, \frac{3}{2}\right)$

$\begin{aligned} & \therefore \ell_1=\sqrt{\frac{1}{36}+\frac{1}{9}+\frac{1}{36}} \\ & \ell_1=\sqrt{\frac{1}{6}}=\ell_2=\ell_3 \\ & \therefore \ell_1^2+\ell_2^2+\ell_3^2=\frac{1}{2} \end{aligned}$

$\begin{aligned} & \mathrm{AB}=5 \\ & \mathrm{AC}=5 \end{aligned}$

$\therefore \mathrm{D}$ is midpoint of $\mathrm{BC}$

$\begin{aligned} & \mathrm{D}\left(\frac{1}{2}, \frac{3}{2}, 3\right) \\ & \therefore l=\sqrt{\left(2-\frac{1}{2}\right)^2+\left(-3-\frac{3}{2}\right)^2+(3-3)^2} \\ & l=\sqrt{\frac{45}{2}} \\ & \therefore 2 l^2=45 \end{aligned}$

$\begin{aligned} & \vec{a} \cdot[(\vec{c} \times \vec{b})-\vec{b}-\vec{c}] \\ & \vec{a} \cdot(\vec{c} \times \vec{b})-\vec{a} \cdot \vec{b}-\vec{a} \cdot \vec{c} \quad \text{..... (i)} \end{aligned}$

$\begin{aligned} & \text { given } \vec{a} \times \vec{c}=\vec{b} \\ & \Rightarrow(\vec{a} \times \vec{c}) \cdot \vec{b}=\vec{b} \cdot \vec{b}=|\vec{b}|^2=27 \\ & \Rightarrow \vec{a} \cdot(\vec{c} \times \vec{b})=[\vec{a} \quad \vec{c} \quad \vec{b}]=(\vec{a} \times \vec{c}) \cdot \vec{b}=27 \quad \text{.... (ii)} \end{aligned}$

$\begin{aligned} & \text { Now } \vec{a} \cdot \vec{b}=3-6+3=0 \quad \text{.... (iii)}\\ & \vec{a} \cdot \vec{c}=3 \quad \text{.... (iv) (given)} \end{aligned}$

$\begin{gathered} \text { By (i), (ii), (iii) & (iv) } \\ 27-0-3=24 \end{gathered}$

Step 1: Given condition for coplanarity

For three vectors to be coplanar, their scalar triple product must be zero. We have the vectors A, B, and C, and we know the given relation between λ and μ:

$A = \lambda \hat{i} - \hat{j} + \hat{k}$

$B = \hat{i} + 2 \hat{j} + \mu \hat{k}$

$C = 3 \hat{i} - 4 \hat{j} + 5 \hat{k}$

Scalar triple product condition:

$[A, B, C] = A \cdot (B \times C) = 0$

Step 2: Calculate the cross product of B and C

$B \times C = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & \mu \\ 3 & -4 & 5 \\ \end{vmatrix}$

$B \times C = (10 + 4 \mu) \hat{i} - (5 - 3 \mu) \hat{j} - 10 \hat{k}$

Step 3: Calculate the scalar triple product and apply the coplanarity condition

$[A, B, C] = A \cdot (B \times C) = \lambda(10 + 4 \mu) - 1(5 - 3 \mu) + 1(-10) = 0$

Step 4: Substitute the given relation between λ and μ Given that:

$\lambda - \mu = 5$

From the coplanarity condition, we have:

$4 \lambda \mu + 10 \lambda - 3 \mu = 5$

Now, substitute λ in terms of μ:

$4(5 + \mu) \mu + 10(5 + \mu) - 3 \mu = 5$

Step 5: Solve for μ and λ

Simplify the equation and solve for μ:

$4 \mu^2 + 27 \mu + 45 = 0$

Factor the equation:

$(4 \mu + 15)(\mu + 3) = 0$

So, the two possible values for μ are:

$\mu = -3, \frac{-15}{4}$

For each value of μ, find the corresponding value of λ using the given relation:

$\lambda = \mu + 5$

So, we get the values for λ:

$\lambda = 2, \frac{5}{4}$

Step 6: Calculate the sum

Now, we need to find the sum:

$\sum\limits_{(\lambda, \mu) \in S} 80\left(\lambda^2 + \mu^2\right)$

Substitute the values of λ and μ, and simplify:

$80\left[(2^2 + (-3)^2) + \left(\frac{5}{4}\right)^2 + \left(\frac{-15}{4}\right)^2\right] = 80\left[13 + \frac{25}{16} + \frac{225}{16}\right]$

$= 80\left[13 + \frac{250}{16}\right] = 10 \times 229$

$= 2290$

So, the correct answer is 2290 (Option D).

Let the position vectors of $A, B, C,$ and $D$ be $\vec{a}, \vec{b}, \vec{c},$ and $\vec{d}$, respectively.

Then the position vector of $E$ is:

$ \vec{E} = \frac{\vec{a} + \vec{c}}{2} $

And the position vector of $F$ is:

$ \vec{F} = \frac{\vec{b} + \vec{d}}{2} $

Now, we are given the equation:

$ (\overrightarrow{AB} - \overrightarrow{BC}) + (\overrightarrow{AD} - \overrightarrow{DC}) = k\overrightarrow{FE} $

We can rewrite this equation using the position vectors:

$ (\vec{b} - \vec{a} - (\vec{c} - \vec{b})) + (\vec{d} - \vec{a} - (\vec{c} - \vec{d})) = k(\vec{E} - \vec{F}) $

Simplifying the equation, we get:

$ (2\vec{b} - 2\vec{a} - 2\vec{c} + 2\vec{d}) = \frac{k}{2}(2\vec{E} - 2\vec{F}) $

Now substitute $\vec{E}$ and $\vec{F}$ expressions we found earlier:

$ (2\vec{b} - 2\vec{a} - 2\vec{c} + 2\vec{d}) = \frac{k}{2}\left(2\left(\frac{\vec{a} + \vec{c}}{2}\right) - 2\left(\frac{\vec{b} + \vec{d}}{2}\right)\right) $

Simplifying the equation:

$ (2\vec{b} - 2\vec{a} - 2\vec{c} + 2\vec{d}) = -\frac{k}{2}(\vec{b} + \vec{d} - \vec{a} - \vec{c}) $

Since both sides of the equation are equal:

$ k = -4 $

$ \begin{aligned} & |\vec{a}|=2 \\\\ & |\vec{b}|=3 \\\\ & \vec{a} \cdot \vec{b}=\frac{\pi}{4} \end{aligned} $

$ \begin{aligned} & |(\vec{a}+2 \vec{b}) \times(2 \vec{a}-3 \vec{b})|^2 \\\\ & = |-3 \vec{a} \times \vec{b}+4 \vec{b} \times \vec{a}|^2 \\\\ & = |-3 \vec{a} \times \vec{b}-4 \vec{a} \times \vec{b}|^2 \\\\ & = |-7 \vec{a} \times \vec{b}|^2 \\\\ & = \left(-7|\vec{a}| \times|\vec{b}| \sin \left(\frac{\pi}{4}\right)\right)^2 \end{aligned} $

= $ 49 \times 4 \times 9 \times \frac{1}{2}=882 $

$ \begin{aligned} & C A+A B=C B \\\\ & \Rightarrow +2 i+4 j+(\delta+3) k=\alpha i+\beta j+\gamma k \\\\ & \Rightarrow \alpha=+2, \beta=4, \gamma=\delta+3 \end{aligned} $

$ \begin{aligned} \text { Area } & =\frac{1}{2}|\overrightarrow{A B} \times \overrightarrow{B C}|=\left| {{1 \over 2}\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr { - 2} & 1 & 3 \cr { + 2} & 4 & \gamma \cr } } \right|} \right|=5 \sqrt{6} \\\\ & =(\gamma-12)^2+(6+2 \gamma)^2+100=(10 \sqrt{6})^2 \\\\ & \Rightarrow 5 \gamma^2=320 \end{aligned} $

$ \begin{aligned} \Rightarrow\gamma^2 =64 \\\\ \Rightarrow\gamma =8 \\\\ \Rightarrow\delta =5 \\\\ \overrightarrow{C B} \cdot \overrightarrow{C A} & =(2 i+4 j+8 k)(4 i+3 j+5 k) \\\\ & =8+12+40 \\\\ & =60 \end{aligned} $

Given that $\vec{d} \times \vec{b} = \vec{c} \times \vec{b}$, we can rewrite this as:

$(\vec{d} - \vec{c}) \times \vec{b} = \vec{0}$

This implies that the vector $\vec{d} - \vec{c}$ is a scalar multiple of $\vec{b}$:

$\vec{d} = \vec{c} + \lambda \vec{b}$

Also, we are given that $\vec{d} \cdot \vec{a} = 24$:

$(\vec{c} + \lambda \vec{b}) \cdot \vec{a} = 24$

Now, we can find the value of $\lambda$ :

$\lambda = \frac{24 - \vec{a} \cdot \vec{c}}{\vec{b} \cdot \vec{a}} = \frac{24 - 6}{9} = 2$

Therefore, we have :

$\vec{d} = \vec{c} + 2(\vec{b}) = 8\hat{i} - 5\hat{j} + 18\hat{k}$

Now, we can find the squared magnitude of $\vec{d}$ :

$|\vec{d}|^2 = 64 + 25 + 324 = 413$

$ \left|\begin{array}{lll} a & 1 & 1 \\\\ 1 & \mathrm{~b} & 1 \\\\ 1 & 1 & \mathrm{c} \end{array}\right|=0 $

$ \mathrm{C}_2 \rightarrow \mathrm{C}_2-\mathrm{C}_1, \mathrm{C}_3 \rightarrow \mathrm{C}_3-\mathrm{C}_1 $

$ \begin{aligned} & \left|\begin{array}{lll} a & 1-a & 1-a \\ 1 & b-1 & 0 \\ 1 & 0 & c-1 \end{array}\right|=0 \\\\ & a(b-1)(c-1)-(1-a)(c-1)+(1-a)(1-b)=0 \\\\ & a(1-b)(1-c)+(1-a)(1-c)+(1-a)(1-b)=0 \end{aligned} $

$ \begin{aligned} & \frac{\mathrm{a}}{1-\mathrm{a}}+\frac{1}{1-\mathrm{b}}+\frac{1}{1-\mathrm{c}}=0 \\\\ & \Rightarrow-1+\frac{1}{1-\mathrm{a}}+\frac{1}{1-\mathrm{b}}+\frac{1}{1-\mathrm{c}}=0 \\\\ & \Rightarrow \frac{1}{1-\mathrm{a}}+\frac{1}{1-\mathrm{b}}+\frac{1}{1-\mathrm{c}}=1 \end{aligned} $

The given vectors are :

$\vec{a} = \lambda \hat{i} + \hat{j} - \hat{k}$

$\vec{b} = 3\hat{i} - \hat{j} + 2\hat{k}$

We are given that $(\vec{a} + \vec{b} + \vec{c}) \times \vec{c} = 0$ which implies $(\vec{a} + \vec{b}) \times \vec{c} = 0$. So, $\vec{c}$ is in the direction of $\vec{a} + \vec{b}$.

Let's denote $\vec{c} = \alpha (\vec{a} + \vec{b})$.

Substituting values for $\vec{a}$ and $\vec{b}$, we get :

$\vec{c} = \alpha((\lambda + 3) \hat{i} + \hat{k})$

From the conditions $\vec{a} \cdot \vec{c} = -17$ and $\vec{b} \cdot \vec{c} = -20$, we can form two equations :

$\alpha \lambda(\lambda + 3) - \alpha = -17 \quad \text{and} \quad \alpha(3\lambda + 9 + 2) = -20.$

By solving the above equations, we find $\alpha = -1$ and $\lambda = 3$.

Substituting these values back into $\vec{c}$ gives $\vec{c} = -6 \hat{i} - \hat{k}$.

Next, we need to find $|\vec{c} \times (\lambda \hat{i} + \hat{j} + \hat{k})|^2$.

This simplifies to :

$|\vec{c} \times (3\hat{i} + \hat{j} + \hat{k})|^2$

Calculate the cross product $\vec{c} \times (3\hat{i} + \hat{j} + \hat{k})$ which gives :

$ =\left|\begin{array}{lll} \hat{i} & \hat{j} & \hat{k} \\ -6 & 0 & -1 \\ 3 & 1 & 1 \end{array}\right|=\hat{\mathrm{i}}+3 \hat{\mathrm{j}}-6 \hat{\mathrm{k}} $

Finally, the square of the magnitude of this vector is:

$|\vec{c} \times (3\hat{i} + \hat{j} + \hat{k})|^2 = (1)^2 + 3^2 + (-6)^2 = 1 + 9 + 36 = 46.$

$ \begin{aligned} & {[\vec{b}-\vec{a} \,\,\,\,\,\vec{c}-\vec{a} \,\,\,\,\,\vec{d}-\vec{a}]=0} \\\\ & (\vec{b}-\vec{a}) \cdot[(\vec{c}-\vec{a}) \times(\vec{d}-\vec{a})]=0 \\\\ & (\vec{b}-\vec{a}) \cdot(\vec{c} \times \vec{d}-\vec{c} \times \vec{a}-\vec{a} \times \vec{d})=0 \\\\ & {[\vec{b}\,\,\,\,\, \vec{c} \,\,\,\,\,\vec{d}]-[\vec{b} \,\,\,\,\,\vec{c} \,\,\,\,\,\vec{a}]-[\vec{b} \,\,\,\,\,\vec{a} \,\,\,\,\,\vec{d}]-[\vec{a} \,\,\,\,\,\vec{c} \,\,\,\,\,\vec{d}]=0} \\\\ & \therefore [\vec{a} \,\,\,\,\,\vec{b} \,\,\,\,\,\vec{c}]=[\vec{b} \,\,\,\,\,\vec{c} \,\,\,\,\,\vec{d}]+[\vec{a} \,\,\,\,\,\vec{b} \,\,\,\,\,\vec{d}]+[\vec{a} \,\,\,\,\,\vec{d} \,\,\,\,\,\vec{c}] \\\\ & \quad=[\vec{d} \,\,\,\,\,\vec{c} \,\,\,\,\,\vec{a}]+[\vec{b} \,\,\,\,\,\vec{d} \,\,\,\,\,\vec{a}]+[\vec{c} \,\,\,\,\,\vec{d} \,\,\,\,\,\vec{b}] \end{aligned} $

We have,

$ \begin{aligned} & 10\left|a_i\right|<1, i=1,2,3 \\\\ & \text { Let } \left|a_1\right| \geq\left|a_2\right| \geq\left|a_3\right| \\\\ & |\vec{a}|=\sqrt{a_1^2+a_2^2+a_3^2} \geq \sqrt{a_1^2} \\\\ & \therefore|\vec{a}| \geq\left|a_1\right| \text { or } \max \left\{\left|a_1\right|,\left|a_2\right|,\left|a_3\right|\right\} \text {. } \end{aligned} $

Hence, (A) is true.

$ \begin{array}{rlrl} & |\vec{a}| =\sqrt{a_1^2+a_2^2+a_3^2} \leq \sqrt{a_1^2+a_1^2+a_1^2} \\\\ & =\sqrt{3}\left|a_1\right| \\\\ \therefore & |\vec{a}|=\sqrt{3}\left|a_1\right|<3\left|a_1\right| \\\\ \therefore & |\vec{a}|<3 \max \left\{\left|a_1\right|,\left|a_2\right|,\left|a_3\right|\right\} \end{array} $

Hence, (B) is also true.

We have, $\vec{a}$ is non-zero vector parallel to the line of intersection of the two planes described by $\hat{\mathbf{i}}+\hat{\mathbf{j}}, \hat{\mathbf{i}}+\hat{\mathbf{k}}$ and $\hat{\mathbf{i}}-\hat{\mathbf{j}}, \hat{\mathbf{j}}-\hat{\mathbf{k}}$.

Let $\mathbf{n}_1$ and $\mathbf{n}_2$ are the normal vector to the plane $\hat{\mathbf{i}}+\hat{\mathbf{j}}, \hat{\mathbf{i}}+\hat{\mathbf{k}}$ and $\hat{\mathbf{i}}-\hat{\mathbf{j}}, \hat{\mathbf{j}}-\hat{\mathbf{k}}$, respectively.

$ \begin{aligned} & n_1=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{array}\right|=\hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}} \\\\ & n_2=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & -1 & 0 \\ 1 & 0 & -1 \end{array}\right|=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}} \\\\ & \vec{a}=\lambda\left|n_1 \times n_2\right| \end{aligned} $

[ $\because \mathbf{a}$ is parallel to line of intersection of both planes]

$ \begin{aligned} & =\lambda\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & -1 & -1 \\ 1 & 1 & 1 \end{array}\right| \\\\ & =\lambda(-2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \end{aligned} $

$\vec{a} \cdot \vec{b} =6 [Given]$

$\lambda(0+4+2) =6 $

$\lambda =1$

$ \begin{aligned} \therefore \vec{a} & =-2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\\\ \cos \theta & =\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}=\frac{6}{\sqrt{4+4} \sqrt{4+4+1}} \\\\ & =\frac{1}{\sqrt{2}} \end{aligned} $

$ \therefore \theta=\frac{\pi}{4} $

$ \begin{aligned} \vec{a} \times \vec{b} & =\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 0 & -2 & 2 \\ 2 & -2 & 1 \end{array}\right| \\\\ & =2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+4 \hat{\mathbf{k}} \\\\ |\vec{a} \times \vec{b}| & =\sqrt{4+16+16}=6 \\\\ \text { Hence, }(\theta,|\vec{a} \times \vec{b}|) & =\left(\frac{\pi}{4}, 6\right) \end{aligned} $

If $\vec{d}$ is $\perp$ to both $\vec{a}$ and $\vec{b}$ then

$ \vec{d}=\lambda(\vec{a} \times \vec{b})=\lambda\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 7 & -1 \\ 3 & 0 & 5 \end{array}\right|=(35 \hat{i}-13 \hat{j}-21 \hat{k}) \lambda $

$ \begin{aligned} & \text { but } \vec{c} \cdot \vec{d}=12 \Rightarrow \lambda(35 \times 1+13 \times 1-21 \times 2)=12 \\\\ & \Rightarrow \lambda(6)=12 \Rightarrow \lambda=2 \\\\ & \vec{\lambda}=2(35 \hat{i}-13 \hat{j}-21 \hat{k}) \end{aligned} $

$ \begin{aligned} & \text { Now, }(-\hat{i}+\hat{j}-\hat{k}) \cdot(\vec{c} \times \vec{d}) \\\\ & =\left|\begin{array}{ccc} -1 & 1 & -1 \\ 1 & -1 & 2 \\ 70 & -26 & -42 \end{array}\right| \\\\ & =-94+182-44=44 \end{aligned} $

1. **Circumcenter $ P $**:

The circumcenter of a triangle is equidistant from the vertices of the triangle. It is the center of the circumcircle, the circle that passes through all three vertices of the triangle.

2. **Orthocenter $ Q $**:

The orthocenter of a triangle is the point of intersection of its three altitudes. An altitude of a triangle is a perpendicular line segment drawn from a vertex to its opposite side (or its extension).

3. **Centroid $ G $**:

The centroid of a triangle is the point of intersection of its medians. A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side. The centroid always divides each median in a 2:1 ratio, with the larger segment being closer to the vertex.

With the above definitions, it's known that the centroid divides the line segment joining the circumcenter and orthocenter in the ratio 2 : 1, meaning :

$ \overrightarrow{PG} = \frac{2}{3} \overrightarrow{PQ} $

$ \overrightarrow{PQ} = 3\overrightarrow{PG} $

The position vector of the centroid $G$ in terms of the vertices $A, B,$ and $C$ is :

$ \overrightarrow{G} = \frac{\overrightarrow{A} + \overrightarrow{B} + \overrightarrow{C}}{3} $

Because $P$ is the circumcenter and is at the origin in this problem:

$ \overrightarrow{PA} = \overrightarrow{A} $

$ \overrightarrow{PB} = \overrightarrow{B} $

$ \overrightarrow{PC} = \overrightarrow{C} $

Substituting these into the equation for $G$ :

$ \overrightarrow{PG} = \frac{\overrightarrow{PA} + \overrightarrow{PB} + \overrightarrow{PC}}{3} $

Now, using the relationship between $PQ$ and $PG$ established earlier :

$ \overrightarrow{PQ} = 3\overrightarrow{PG} = \overrightarrow{PA} + \overrightarrow{PB} + \overrightarrow{PC} $

Given, the position vector of point P is : $ \overrightarrow{OP} = -\widehat{i} - 2\widehat{j} + 3\widehat{k} $

Position vectors of points A, B, and C are :

$ \overrightarrow{OA} = -2\widehat{i} + \widehat{j} - 3\widehat{k} $

$ \overrightarrow{OB} = 2\widehat{i} + 4\widehat{j} - 2\widehat{k} $

$ \overrightarrow{OC} = -4\widehat{i} + 2\widehat{j} - \widehat{k} $

Now, vectors $ \overrightarrow{AB} $ and $ \overrightarrow{AC} $ can be calculated as :

$ \overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} $

$ = (2 + 2)\widehat{i} + (4 - 1)\widehat{j} - (-2 + 3)\widehat{k} $

$ = 4\widehat{i} + 3\widehat{j} - \widehat{k} $

$ \overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} $

$ = (-4 + 2)\widehat{i} + (2 - 1)\widehat{j} - (-1 + 3)\widehat{k} $

$ = -2\widehat{i} + \widehat{j} - 2\widehat{k} $

$ \begin{aligned} \text { Now, } \overrightarrow{A B} \times \overrightarrow{A C} & =\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 4 & 3 & 1 \\ -2 & 1 & 2 \end{array}\right| \\\\ & =\hat{\mathbf{i}}(5)+\hat{\mathbf{j}}(-2-8)+\hat{\mathbf{k}}(4+6) \\\\ & =5 \hat{\mathbf{i}}-10 \hat{\mathbf{j}}+10 \hat{\mathbf{k}} \\\\ &\overrightarrow{O P} =-\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}} \end{aligned} $

To find the projection of $ \overrightarrow{OP} $ onto $ \overrightarrow{AB} \times \overrightarrow{AC} $, we need to find the dot product between $ \overrightarrow{OP} $ and the normalized vector $ \overrightarrow{AB} \times \overrightarrow{AC} $.

First, find the magnitude of $ \overrightarrow{AB} \times \overrightarrow{AC} $ :

$ |\overrightarrow{AB} \times \overrightarrow{AC}| = \sqrt{5^2 + (-10)^2 + 10^2} $

$ = \sqrt{25 + 100 + 100} $

$ = \sqrt{225} $

$ = 15 $

The projection of vector $ \overrightarrow{OP} $ onto a vector perpendicular to both $ \overrightarrow{AB} $ and $ \overrightarrow{AC} $ (which is $ \overrightarrow{AB} \times \overrightarrow{AC}$) is given by :

$ \frac{\overrightarrow{OP} \cdot (\overrightarrow{AB} \times \overrightarrow{AC})}{| \overrightarrow{AB} \times \overrightarrow{AC} |} $

= $ \frac{(-\hat{i} - 2\hat{j} + 3\hat{k}) \cdot (5\hat{i} - 10\hat{j} + 10\hat{k})}{\sqrt{25 + 100 + 100}} $

$ = \frac{-5 + 20 + 30}{15} $

$ = \frac{45}{15} = 3 $

An arc $P Q$ of a circle subtends a right angle at its centre $O$. The mid-point of an $\operatorname{arc} P Q$ is $R$. So, $P R=R Q$

Also given that, $\overrightarrow {OP} = \overrightarrow u ,\overrightarrow {OR} = \overrightarrow v $, and $\overrightarrow {OQ} = \alpha \overrightarrow u + \beta \overrightarrow v $



Let $\overrightarrow {OP} = \overrightarrow u$

$ \begin{aligned} & \overrightarrow{OQ}=\overrightarrow{q}=\alpha \overrightarrow{u}+\beta \overrightarrow{v} \\\\ & \overrightarrow{O R}=\overrightarrow{v} \end{aligned} $

Clearly, $\overrightarrow{q}=\hat{\mathbf{j}}$

$ \overrightarrow{u}=\hat{\mathbf{i}}, \overrightarrow{v}=\frac{\hat{\mathbf{i}}+\hat{\mathbf{j}}}{\sqrt{2}} $

Now, given, $\overrightarrow{q}=\alpha \overrightarrow{u}+\beta \overrightarrow{v}$

$ \begin{array}{ll} \Rightarrow & \hat{\mathbf{j}}=\alpha(\hat{\mathbf{i}})+\beta\left(\frac{\hat{\mathbf{i}}+\hat{\mathbf{j}}}{\sqrt{2}}\right) \\\\ \Rightarrow & \hat{\mathbf{j}}=\mathbf{i}\left(\alpha+\frac{\beta}{\sqrt{2}}\right)+\frac{\beta}{\sqrt{2}} \hat{\mathbf{j}} \end{array} $

On comparing both sides, we get

$ \begin{array}{rlrl} \alpha+\frac{\beta}{\sqrt{2}} =0 \text { and } \frac{\beta}{\sqrt{2}}=1 \\\\ \Rightarrow \alpha =-\frac{\beta}{\sqrt{2}} \quad \therefore \beta =\sqrt{2} \\\\ \Rightarrow \alpha =-1, \beta^2=2 \end{array} $

Now, $\alpha$ and $\beta^2$ are roots of the quadratic equation is $x^2-x-2=0$

Since, $\vec{u}_1, \vec{u}_2, \vec{u}_3$ are coplanar.

So, $\left[\begin{array}{lll}\vec{u}_1 & \vec{u}_2 & \vec{u}_3\end{array}\right]=0$

$ \begin{aligned} & \Rightarrow\left|\begin{array}{lll} 1 & 1 & a \\ 1 & b & 1 \\ c & 1 & 1 \end{array}\right|=0 \\\\ & \Rightarrow 1(b-1)-1(1-c)+a(1-b c)=0 \\\\ & \Rightarrow b-1-1+c+a-a b c=0 \\\\ & \Rightarrow a+b+c-2=a b c ........... (i) \end{aligned} $

Also, $\left[\begin{array}{lll}\vec{v}_1 & \vec{v}_2 & \vec{v}_3\end{array}\right]=0$

$ \begin{aligned} & \Rightarrow\left|\begin{array}{ccc} a+b & c & c \\ a & b+c & a \\ b & b & c+a \end{array}\right|=0 \\ & \Rightarrow(a+b)\left[b c+b a+c^2+c a-a b\right]-c\left[a c+a^2-a b\right] \\\\ & \quad+c\left[a b-b^2-b c\right]=0 \end{aligned} $

$ \begin{aligned} & \Rightarrow a b c+a c^2+a^2 c+b^2 c+b c^2+a b c-a c^2-a^2 c \\\\ & +a b c+a b c-b^2 c-b c^2=0 \\\\ & \Rightarrow 4 a b c=0 \Rightarrow a b c=0 \\\\ & \begin{array}{lr} \text { So, } a+b+c-2=0 [from (i)]\\\\ \Rightarrow a+b+c=2 \end{array} \\\\ & \Rightarrow 6(a+b+c)=12 \end{aligned} $