Vector Algebra

• Point $P$ : $(43, \alpha, \beta)$ with the constraint $\beta<0$.

• Target Line ( $L_1$ ) : Given by $\vec{r}=4 \hat{i}-\hat{k}+\mu(2 \hat{i}+3 \hat{k})$.
  
  A general point $Q$ on this line is $(4+2 \mu, 0,-1+3 \mu)$.

• Direction of measurement : The distance is measured along a line with direction ratios $(3,-1,0)$.

Find Point $Q$ on the Target Line

Since the distance is measured along the direction $d=(3,-1,0)$, the vector $P Q$ must be parallel to $d$.

$ \begin{gathered} P Q=Q-P=(4+2 \mu-43, \quad 0-\alpha, \quad-1+3 \mu-\beta) \\ P Q=(2 \mu-39, \quad-\alpha, \quad 3 \mu-\beta-1) \\ P Q=(2 \mu-39, \quad-\alpha, \quad 3 \mu-\beta-1) \end{gathered} $

Because $P Q$ is parallel to $(3,-1,0)$, their components must be proportional :

1. 1.$\frac{2 \mu-39}{3}=\frac{-\alpha}{-1} \Longrightarrow 2 \mu-39=3 \alpha \Longrightarrow 2 \mu=3 \alpha+39$

2. 2.$\frac{3 \mu-\beta-1}{0}$ suggests the numerator must be zero for a finite ratio to exist: $3 \mu-\beta-1= 0 \Longrightarrow \beta=3 \mu-1$

Calculate the Distance $P Q$

The distance $P Q$ is given as $13 \sqrt{10}$. Using the distance formula on the proportional components where the ratio is $k$ :

$ \begin{gathered} P Q=k(3,-1,0)=(3 k,-k, 0) \\ |P Q|=\sqrt{(3 k)^2+(-k)^2+0^2}=\sqrt{10 k^2}=|k| \sqrt{10} \end{gathered} $

Given $|P Q|=13 \sqrt{10}$, we find $|k|=13$, so $k= \pm 13$.

Solve for $\alpha, \mu$, and $\beta$

From our proportionality equations :

• $-\alpha=-k \Longrightarrow \alpha=k$

• $2 \mu-39=3 k$

Case 1 : $k=13$

• $\alpha=13$

• $2 \mu-39=3(13)=39 \Longrightarrow 2 \mu=78 \Longrightarrow \mu=39$

• $\beta=3 \mu-1=3(39)-1=116$

• Constraint Check : $\beta<0$ is False. (Reject this case)

Case 2 : $k=-13$

• $\alpha=-13$

• $2 \mu-39=3(-13)=-39 \Longrightarrow 2 \mu=0 \Longrightarrow \mu=0$

• $\beta=3(0)-1=-1$

• Constraint Check : $\beta<0$ is True. (Accept this case)

$\alpha^2+\beta^2=(-13)^2+(-1)^2=169+1=170$

The value of $\alpha^2+\beta^2$ is 170 .

We are given that $|\overrightarrow{P R}|=9$ Using the distance formula for the vector $P R=a \hat{i}+b \hat{j}-4 \hat{k}$ :

$ \sqrt{a^2+b^2+(-4)^2}=9 $

$\Rightarrow $ $a^2+b^2+16=81$

$\Rightarrow $ $ a^2+b^2=65 $

The problem states that point $S$ lies on $Q R$ and is equidistant from lines $P Q$ and $P R$. In a triangle, the locus of points equidistant from two sides is the angle bisector. Therefore, $\overrightarrow{P S}$ is the internal angle bisector of $\angle Q P R$.

For an angle bisector $\overrightarrow{P S}$, it must be collinear with the sum of the unit vectors along the adjacent sides :

$ \overrightarrow{P S}=\lambda\left(\frac{\overrightarrow{P Q}}{|\overrightarrow{P Q}|}+\frac{\overrightarrow{P R}}{|\overrightarrow{P R}|}\right) $

where $\lambda$ is a scalar constant.

First, calculate $|\overrightarrow{P Q}|$ :

$ |\overrightarrow{P Q}|=\sqrt{(-2)^2+(-1)^2+2^2}=\sqrt{4+1+4}=3 $

Now, set up the bisector equation :

$ \overrightarrow{P S}=\lambda\left(\frac{-2 \hat{i}-\hat{j}+2 \hat{k}}{3}+\frac{a \hat{i}+b \hat{j}-4 \hat{k}}{9}\right) $

Multiply the first term by $\frac{3}{3}$ to get a common denominator:

$ \overrightarrow{P S}=\frac{\lambda}{9}((-6+a) \hat{i}+(-3+b) \hat{j}+(6-4) \hat{k}) $

$\Rightarrow $ $ \overrightarrow{P S}=\frac{\lambda}{9}((a-6) \hat{i}+(b-3) \hat{j}+2 \hat{k}) $

We are given

$ \overrightarrow{P S}=\hat{i}-7 \hat{j}+2 \hat{k} $

Comparing the $\hat{k}$-components:

$\Rightarrow $ $ \frac{\lambda}{9}(2)=2 \Rightarrow \lambda=9 $

Substitute $\lambda=9$ into other components:

For $\hat{i}$ :

$ a-6=1 \Rightarrow a=7 $

For $\hat{j}$ :

$ b-3=-7 \Rightarrow b=-4 $

Check against the constraint :

$ a^2+b^2=7^2+(-4)^2=49+16=65 $

The values are consistent. value of $3 a-4 b$ :

$ 3 a-4 b=3(7)-4(-4) $ $=21+16=\mathbf{3 7}$

$ \begin{aligned} & \vec{a}^{\wedge} \hat{k} \in\left(\frac{\pi}{6}, \frac{\pi}{2}\right) \\ & \frac{\lambda}{\sqrt{\left(2+1+\lambda^2\right)}} \in\left(0, \frac{\sqrt{3}}{2}\right) \Rightarrow 0<\frac{\lambda}{\sqrt{3+\lambda^2}}<\frac{\sqrt{3}}{2} \\ & \lambda>0 \& \lambda \in(-3,3) \Rightarrow \lambda \in(0,3) \ldots \ldots \ldots .(1) \end{aligned} $

$\vec{a}^{\wedge} \vec{b}$ is obtuse

$ \begin{aligned} & \vec{a} \cdot \vec{b}<0 \Rightarrow-\sqrt{2} \lambda^2-4 \sqrt{2}+4 \sqrt{2} \lambda<0 \\ & \Rightarrow \lambda^2-4 \lambda+4>0 \Rightarrow(\lambda-2)^2>0 \\ & \Rightarrow \lambda \in R-\{2\} \ldots \ldots .(2) \end{aligned} $

(1) \& (2) ⇒ $\lambda \in(0,3)-\{2\} \Rightarrow \alpha+\beta+\gamma=5$

$\begin{aligned} &\text { Assuming Vertex } A \text { to be origin }\\ &\begin{aligned} & \vec{A}=\vec{a}_1=\overrightarrow{0} \\ & \vec{B}=\vec{a}_1+\vec{u}=\vec{u}=2 \hat{i}-\hat{j}+\hat{k} \\ & \vec{C}=\vec{a}_1+\vec{v}=\vec{v}=3 \hat{i}-4 \hat{j}-4 \hat{k} \end{aligned} \end{aligned}$

One solving

$\vec{A}=\overrightarrow{0}, \vec{B}=2 \hat{i}-\hat{j}+\hat{k}$ and $\vec{C}=3 \hat{i}-4 \hat{j}-4 \hat{k}$, are the position vector of vertices $A B$ and $C$ respectively.

$\begin{aligned} & \vec{G}=\frac{1}{3}(\vec{A}+\vec{B}+\vec{C})=\frac{1}{3}(\overrightarrow{0}+\vec{B}+\vec{C})=\frac{1}{3}(\vec{B}+\vec{C}) \\ & \Rightarrow \vec{G}=\frac{5}{3} \hat{i}-\frac{5}{3} \hat{j}-\hat{k} \\ & \overrightarrow{A G}=\vec{G}-\vec{A}=\vec{G} \\ & |\overrightarrow{A G}|^2=\left(\frac{5}{3}\right)^2+\left(\frac{5}{3}\right)^2+(1)^2=\frac{25}{9}+\frac{25}{9}+1=\frac{50}{9}+1=\frac{59}{9} \\ & \overrightarrow{B C}=\vec{G}-\vec{B} \\ & \vec{B}=2 \hat{i}-\hat{j}+\hat{k} \\ & |\overrightarrow{B G}|^2=\left(\frac{1}{3}\right)^3+\left(\frac{2}{3}\right)^2+4=\frac{1}{9}+\frac{4}{9}+4=\frac{5}{9}+4=\frac{41}{9} \\ & \overrightarrow{C G}=\vec{G}-\vec{C} \\ & \vec{C}=3 \hat{i}-4 \hat{j}-4 \hat{k} \\ & \left.\overrightarrow{C G}\right|^2=\left(\frac{4}{3}\right)^2+\left(\frac{7}{3}\right)^2+9=\frac{16}{9}+\frac{49}{9}+9=\frac{65}{9}+9=\frac{65}{9}+\frac{81}{9}=\frac{146}{9} \end{aligned}$

$6\left(|\overrightarrow{A G}|^2+|\overrightarrow{B G}|^2+|\overrightarrow{C G}|^2\right)=6 \cdot\left(\frac{59}{9}+\frac{41}{9}+\frac{146}{9}\right)=6 \cdot \frac{246}{9}=164$

Given Conditions :

The condition $\vec{b} \times \vec{d} = \vec{c} \times \vec{d}$ implies that $(\vec{b} - \vec{c}) \times \vec{d} = 0$.

This indicates that $\vec{d}$ is parallel to $\vec{b} - \vec{c}$.

Expressing $\vec{b} - \vec{c}$:

Calculate $\vec{b} - \vec{c} = (3 \hat{i} - 3 \hat{j} + 3 \hat{k}) - (2 \hat{i} - \hat{j} + 2 \hat{k}) = \hat{i} - 2 \hat{j} + \hat{k}$.

Express $\vec{d}$:

Since $\vec{d}$ is parallel to $\vec{b} - \vec{c}$, let $\vec{d} = \lambda(\hat{i} - 2 \hat{j} + \hat{k})$.

Using the condition $\vec{a} \cdot \vec{d} = 4$:

Compute $\vec{a} \cdot \vec{d} = (\hat{i} + 2 \hat{j} + \hat{k}) \cdot (\lambda(\hat{i} - 2 \hat{j} + \hat{k}))$.

This simplifies to $\lambda(1 \cdot 1 + 2 \cdot (-2) + 1 \cdot 1) = \lambda(1 - 4 + 1) = -2\lambda$.

Set $-2\lambda = 4$, which gives $\lambda = -2$.

Determine $\vec{d}$ using $\lambda$:

Therefore, $\vec{d} = -2(\hat{i} - 2 \hat{j} + \hat{k}) = -2 \hat{i} + 4 \hat{j} - 2 \hat{k}$.

Calculate $\vec{a} \times \vec{d}$:

Use the determinant form for the cross product:

$ \vec{a} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 1 \\ -2 & 4 & -2 \end{vmatrix} $

Calculate the determinant:

For $\hat{i}$: $(2 \times -2) - (1 \times 4) = -4 - 4 = -8$

For $\hat{j}$: $(-(1 \times -2) - (1 \times -2)) = 2 - 2 = 0$

For $\hat{k}$: $(1 \times 4) - (2 \times -2) = 4 + 4 = 8$

Thus, $\vec{a} \times \vec{d} = -8 \hat{i} + 0 \hat{j} + 8 \hat{k}$.

Calculate the magnitude squared:

Find $|\vec{a} \times \vec{d}|^2 = (-8)^2 + 0^2 + 8^2 = 64 + 0 + 64 = 128$.

Therefore, $|(\vec{a} \times \vec{d})|^2 = 128$.

Given the vectors $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = 3 \hat{i} + 2 \hat{j} - \hat{k}$, along with $\vec{c} = \lambda \hat{j} + \mu \hat{k}$, and $\hat{d}$ being a unit vector such that $\vec{a} \times \hat{d} = \vec{b} \times \hat{d}$ and $\vec{c} \cdot \hat{d} = 1$, we proceed as follows:

Determine $\hat{d}$:

Since $\vec{a} \times \hat{d} = \vec{b} \times \hat{d}$, we have:

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

Thus, $\hat{d}$ is parallel to $\vec{a} - \vec{b}$. Calculate:

$ \vec{a} - \vec{b} = (-2 \hat{i} + \hat{j} + 2 \hat{k}) $

Therefore, we can express $\hat{d}$ as:

$ \hat{d} = \frac{-2}{3} \hat{i} - \frac{1}{3} \hat{j} + \frac{2}{3} \hat{k} $

Solve for $\lambda$ and $\mu$:

Using the condition $\vec{c} \cdot \hat{d} = 1$:

$ \frac{-\lambda}{3} + \frac{2 \mu}{3} = 1 $

Simplify to:

$ -\lambda + 2\mu = 3 \quad \text{...(i)} $

Since $\vec{c}$ is perpendicular to $\vec{a}$:

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

Which gives:

$ \lambda + \mu = 0 \quad \Rightarrow \lambda = -\mu $

Substitute $\lambda = -\mu$ into equation (i):

$ \mu + 2\mu = 3 \quad \Rightarrow 3\mu = 3 \quad \Rightarrow \mu = 1 $

Therefore, $\lambda = -1$.

Find $|3 \lambda \hat{d} + \mu \vec{c}|^2$:

Calculate:

$ |3 \lambda \hat{d} + \mu \vec{c}|^2 = 9 \lambda^2 |\hat{d}|^2 + \mu^2 |\vec{c}|^2 + 2 \cdot 3 \cdot \lambda \cdot \mu \cdot \vec{c} \cdot \vec{d} $

Substituting the determined values:

$ |\hat{d}|^2 = 1 $ (since $\hat{d}$ is a unit vector)

$ \vec{c} \cdot \vec{d} = 1 $

Thus:

$ = 9(-1)^2 \cdot 1 + 1^2 \cdot (1^2 + 1^2) + 2 \cdot 3 \cdot (-1) \cdot 1 \cdot 1 $

$ = 9 + 2 - 6 $

$ = 5 $

Hence, $|3 \lambda \hat{d} + \mu \vec{c}|^2 = 5$.

$\begin{aligned} & \overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{~b}}=2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{~d}}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}} \\ & =-\hat{\mathrm{i}}+\hat{\mathrm{j}} \\ & |\overrightarrow{\mathrm{c}}-2 \overrightarrow{\mathrm{a}}|^2=8 \\ & |\mathrm{c}|^2+4|\mathrm{a}|^2-4(\mathrm{a} \cdot \mathrm{c})=8 \\ & \mathrm{c}^2+12-4 \mathrm{c}=8 \\ & \mathrm{c}^2-4 \mathrm{c}+4=0 \\ & |\mathrm{c}|=2 \\ & \overrightarrow{\mathrm{~d}}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}} \\ & \overrightarrow{\mathrm{~d}} \times \overrightarrow{\mathrm{c}}=(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})^2 \times \overrightarrow{\mathrm{c}} \\ & \left(|\mathrm{~d}||\mathrm{c}| \sin \frac{\pi}{4}\right)^2=((\mathrm{a} \cdot \mathrm{c}) \cdot \mathrm{b}-(\mathrm{b} \cdot \mathrm{c}) \cdot \mathrm{a})^2 \\ & 4=4 \mathrm{~b}^2+(\mathrm{b} \cdot \mathrm{c})^2 2(\mathrm{a})^2-2(\mathrm{~b} \cdot \mathrm{c})(\mathrm{a} \cdot \mathrm{~b}) \end{aligned}$

$\begin{aligned} &\text { Let } \mathrm{b} . \mathrm{c}=\mathrm{x}\\ &\begin{aligned} & 4=36+3 x^2-20 x \\ & 3 x^2-20 x+32=0 \\ & 3 x^2-12 x-8 x+32=0 \\ & x=\frac{8}{3}, 4 \\ & \text { b.c }=\frac{8}{3}, 4 \\ & \text { b.c }=\frac{8}{3} \\ & \text { Now }|10-3 \mathrm{~b} . \mathrm{c}|+|\mathrm{d} \times \mathrm{c}|^2 \\ & |10-8|+(2)^2 \\ & \Rightarrow 6 \text { Ans. } \end{aligned} \end{aligned}$

To find the projection vector $\vec{c}$ of $\vec{b} = \lambda \hat{i} + 4 \hat{k}$ (where $\lambda > 0$) onto vector $\vec{a} = \hat{i} + 2 \hat{j} + 2 \hat{k}$, we use the formula for the projection of a vector:

$ \vec{c} = \left(\frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2}\right) \vec{a} $

Calculate the dot product $\vec{b} \cdot \vec{a}$:

$ \vec{b} \cdot \vec{a} = (\lambda \hat{i} + 4 \hat{k}) \cdot (\hat{i} + 2 \hat{j} + 2 \hat{k}) = \lambda \cdot 1 + 4 \cdot 2 = \lambda + 8 $

Calculate the magnitude of $\vec{a}$:

$ |\vec{a}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{9} = 3 $

Now substitute to find $\vec{c}$:

$ \vec{c} = \left(\frac{\lambda + 8}{9}\right)(\hat{i} + 2 \hat{j} + 2 \hat{k}) $

We know that the magnitude $|\vec{a} + \vec{c}| = 7$. Thus, substituting $\vec{c}$ in this equation, we resolve it to find $\lambda$:

$ |\vec{a} + \vec{c}| = 7 \Rightarrow (\lambda = 4) $

This indicates that $\lambda$ has a value of 4.

Next, calculating the area of the parallelogram formed by vectors $\vec{b}$ and $\vec{c}$ involves finding the cross product $|\vec{b} \times \vec{c}|$:

$ \vec{b} = 4\hat{i} + 4\hat{k}, \quad \vec{c} = \left(\frac{12}{9}\right)(\hat{i} + 2\hat{j} + 2\hat{k}) = \left(\frac{4}{3}\right)\hat{i} + \left(\frac{8}{3}\right)\hat{j} + \left(\frac{8}{3}\right)\hat{k} $

The cross product is:

$ \vec{b} \times \vec{c} = \left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ \frac{4}{3} & \frac{8}{3} & \frac{8}{3} \\ 4 & 0 & 4 \end{array}\right| $

Solving the determinant:

$ \vec{b} \times \vec{c} = \left( \left( \frac{8}{3} \times 4 - 0 \times \frac{8}{3} \right)\hat{i} - \left(\frac{4}{3} \times 4 - 4 \times \frac{8}{3} \right)\hat{j} + \left( \frac{4}{3} \times 0 - \frac{8}{3} \times 4 \right)\hat{k} \right) $

$ = \left(\frac{32}{3}\hat{i} + 0\hat{j} + (-\frac{32}{3})\hat{k}\right) $

Then, the magnitude is calculated as:

$ |\vec{b} \times \vec{c}| = \sqrt{\left(\frac{32}{3}\right)^2 + 0 + \left(-\frac{32}{3}\right)^2} = \sqrt{\left(\frac{32}{3}\right)^2 + \left(\frac{32}{3}\right)^2} = \sqrt{\frac{2048}{9}} $

$ = \frac{\sqrt{2048}}{3} = \frac{16}{3} \times 3 \approx 16 $

Therefore, the area of the parallelogram formed by $\vec{b}$ and $\vec{c}$ is 16.

$\begin{aligned} & \vec{a}=9 \hat{i}-13 \hat{j}+25 \hat{k} \\ & \vec{b}=3 \hat{i}+7 \hat{j}-13 \hat{k} \\ & \vec{c}=17 \hat{i}-2 \hat{j}+\hat{k} \\ & \vec{r} \times \vec{a}=(\vec{b}+\vec{c}) \times \vec{a} \\ & (\vec{r}-(\vec{b}+\vec{c})) \times \vec{a}=0 \\ & \Rightarrow \vec{r}=(\vec{b}+\vec{c})+\lambda \vec{a} \\ & \vec{r}=(20 \hat{i}+5 \hat{j}-12 \hat{k})+\lambda(9 \hat{i}-13 \hat{j}+25 \hat{k}) \\ & =(20+9 \lambda) \hat{i}+(5-13 \lambda) \hat{j}+(25 \lambda-12) \hat{k} \end{aligned}$

Now $\vec{r} \cdot(\vec{b}-\vec{c})=0$

$\vec{r} \cdot(-14 \hat{i}+9 \hat{j}-14 \hat{k})=0$

Now

$\begin{aligned} & -14(20+9 \lambda)+9(5-13 \lambda)-14(25 \lambda-12)=0 \\ & -593 \lambda-67=0 \\ & \lambda=-\frac{67}{593} \\ & \therefore \vec{r}=(\vec{b}+\vec{c})-\frac{67}{593} \vec{a} \\ & \frac{|593 \vec{r}+67 \vec{a}|^2}{|593|^2}=|\vec{b}+\vec{c}|^2=|20 \hat{i}+5 \hat{j}-12 \hat{k}|^2 \\ & =569 \end{aligned}$

Let $\hat{i}+8 \hat{j}+13 \hat{k}=\vec{u}$

Given $\vec{a} \times(\vec{b}+\vec{c})+\vec{b} \times \vec{c}=\vec{u}$

$\begin{gathered} \Rightarrow \quad \vec{a} \times \vec{b}+\vec{a} \times \vec{c}+\vec{b} \times \vec{c}=\vec{u} \\ (\vec{a}+\vec{b}) \times c=\vec{u}-\vec{a} \times \vec{b} \end{gathered}$

Taking cross product with $\vec{a}$ on both sides

$\vec{a} \times((\vec{a}+\vec{b}) \times \vec{c})=\vec{a} \times(\vec{u}-\vec{a} \times \vec{b})$

$\Rightarrow \quad \vec{c} \cdot\left(\vec{a}^2+\vec{a} \cdot \vec{b}\right)=13(\vec{a}+\vec{b})-\vec{a} \times \vec{u} +(\vec{a} \cdot \vec{b}) \cdot \vec{a}-\vec{a}^2 \vec{b}\quad$ $\{\because \vec{a} . \vec{c}=13\}$

Putting the values, $\vec{c}=(-1,-1,3)$

$\vec{b}.\vec{c}=-22$

$\Rightarrow 24-\vec{b} \cdot \vec{c}=46$

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

$\begin{aligned} \Rightarrow \quad & \vec{b} \times \vec{c}+2(\vec{a} \times \vec{c})=0 \\ & (\vec{b}+2 \vec{a}) \times \vec{c}=0 \\ & \vec{c}=\lambda(\vec{b}+2 \vec{a}) \\ & \vec{c} \cdot \vec{a}=130 \Rightarrow \lambda=1 \\ & \vec{c}=4 \hat{i}-7 \hat{j}+15 \hat{k} \\ & \vec{b} . \vec{c}=30 \end{aligned}$

Area of triangle $A B C=15 \sqrt{2}$

$\begin{aligned} & \Rightarrow \frac{1}{2}|\overline{A B} \times \overline{A C}|=15 \sqrt{2} \quad \text{.... (i)}\\ & \quad \overline{A B} \times \overline{A C}\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -7 \\ 6 & d & -2 \end{array}\right| \\ & =(7 d-4) \hat{i}-40 \hat{j}+(d-12) \hat{k} \quad \text{... (ii)} \end{aligned}$

From (i) and (ii) $5 d^2-8 d-4=0$

$\Rightarrow d=\frac{-}{5}$ (Rejected) or $d=2$

Also, $\overline{A B}+\overline{B C}=\overline{A C}$

$\begin{aligned} & \Rightarrow a+1=6 \Rightarrow a=5 \\ & b+2=d \Rightarrow b=0 \end{aligned}$

and $c-7=-2 \Rightarrow c=5$

$|\overline{A B}|=\sqrt{54},|\overline{A C}|=\sqrt{44},|\overline{B C}|=\sqrt{50}$

Largest side has length of $\sqrt{54}$ units

$\begin{aligned} & (\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}=2(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})+24 \hat{\mathrm{j}}-6 \hat{\mathrm{k}} \\ & (5 \hat{\mathrm{i}}+\hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \times \overrightarrow{\mathrm{c}}=2(7 \hat{\mathrm{i}}-7 \hat{\mathrm{j}}-7 \hat{\mathrm{k}})+24 \hat{\mathrm{j}}-6 \hat{\mathrm{k}} \\ & \left|\begin{array}{lrr} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 5 & 1 & 4 \\ x & y & \mathrm{z} \end{array}\right|=14 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}-20 \hat{\mathrm{k}} \\ & \Rightarrow \hat{\mathrm{i}}(\mathrm{z}-4 \mathrm{y})-\hat{\mathrm{j}}(5 \mathrm{z}-4 \mathrm{x})+\hat{\mathrm{k}}(5 \mathrm{y}-\mathrm{x})=14 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}-20 \hat{\mathrm{k}} \\ & \mathrm{z}-4 \mathrm{y}=14,4 \mathrm{x}-5 \mathrm{z}=10,5 \mathrm{y}-\mathrm{x}=-20 \\ & (\mathrm{a}-\mathrm{b}+\mathrm{i}) \cdot \overrightarrow{\mathrm{c}}=-3 \\ & (2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}) \cdot \overrightarrow{\mathrm{c}}=-3 \\ & 2 x+3 y-2 z=-3 \\ & \therefore x=5, y=-3, z=2 \\ & |\overrightarrow{\mathrm{c}}|^2=25+9+4=38 \end{aligned}$

$\begin{aligned} & \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=(2 \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \cdot \overrightarrow{\mathrm{b}}-3|\mathrm{b}|^2 \\ & |\mathrm{~b}||c| \cos \alpha=-3|\mathrm{~b}|^2 \\ & |\mathrm{c}| \cos \alpha=-12 \text {, as }|\mathrm{b}|=4 \\ & \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=2 \\ & \cos \theta=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{3} \\ & |c|^2=|(2 \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})-3 \overrightarrow{\mathrm{b}}|^2 \\ & =64 \times \frac{3}{4}+144=192 \\ & |c|^2 \cos ^2 \alpha=144 \\ & 192 \cos ^2 \alpha=144 \\ & 192 \sin ^2 \alpha=48 \end{aligned}$

$\begin{aligned} & \cos \theta=\frac{(\alpha \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \cdot(\alpha \hat{\mathrm{i}}+2 \alpha \hat{\mathrm{j}}-2 \hat{\mathrm{k}})}{\sqrt{\alpha^2+4+4} \sqrt{\alpha^2+4 \alpha^2+4}} \\ & \cos \theta=\frac{\alpha^2-4 \alpha-4}{\sqrt{\alpha^2+8} \sqrt{5 \alpha^2+4}} \\ & \Rightarrow \alpha^2-4 \alpha-4>0 \quad \Rightarrow(\alpha-2)^2>8 \\ & \Rightarrow \alpha^2-4 \alpha+4>8 \quad \alpha-2<-2 \sqrt{2} \\ & \Rightarrow \alpha-2>2 \sqrt{2} \text { or } \alpha-2<2 \sqrt{2} \\ & \alpha>2+2 \sqrt{2} \text { or } \alpha<2-2 \sqrt{2} \\ & \alpha \in(-\infty,-0.82) \cup(4.82, \infty) \end{aligned}$

Least positive integral value of $\alpha \Rightarrow 5$

Given that $\vec{a} = \vec{b} \times \vec{c}$, we can find the magnitudes of $\vec{a}$ and $\vec{c}$:

$|\vec{a}| = \sqrt{3^2 + 1^2 + (-1)^2} = \sqrt{11}$

$|\vec{c}| = \sqrt{2^2 + (-3)^2 + 3^2} = \sqrt{22}$

We know that the magnitude of the cross product of two vectors is equal to the product of the magnitudes of the vectors and the sine of the angle between them :

$|\vec{a}| = |\vec{b} \times \vec{c}| = |\vec{b}||\vec{c}|\sin\theta$

Plugging in the known values :

$\sqrt{11} = \sqrt{50}\sqrt{22}\sin\theta$

Solving for the sine of the angle between the vectors :

$\sin\theta = \frac{1}{10}$

Now we can find $|\vec{b} + \vec{c}|^2$ using the formula :

$|\vec{b} + \vec{c}|^2 = |\vec{b}|^2 + |\vec{c}|^2 + 2\vec{b} \cdot \vec{c}$

We have the dot product $\vec{b} \cdot \vec{c} = |\vec{b}||\vec{c}|\cos\theta$, and we can use the relationship between sine and cosine: $\cos\theta = \sqrt{1 - \sin^2\theta} = \frac{\sqrt{99}}{10}$.

Substitute the values into the formula :

$|\vec{b} + \vec{c}|^2 = 50 + 22 + 2\sqrt{50}\sqrt{22}\frac{\sqrt{99}}{10} = 72 + 66$

Finally, we need to find the absolute value of the difference :

$|72 - |\vec{b} + \vec{c}|^2| = |72 - (72 + 66)| = 66$

Given,

$\overrightarrow a \times \overrightarrow b = 4\,.\,\overrightarrow c $ ..... (i)

$\overrightarrow b \times \overrightarrow c = 9\,.\,\overrightarrow a $ ..... (ii)

$\overrightarrow c \times \overrightarrow a = \alpha \,.\,\overrightarrow b $ .... (iii)

Taking dot products with $\overrightarrow c ,\overrightarrow a ,\overrightarrow b $ we get

$\overrightarrow a \,.\,\overrightarrow b = \overrightarrow b \,.\,\overrightarrow c = \overrightarrow c \,.\,\overrightarrow a = 0$

Hence,

(i) $ \Rightarrow |\overrightarrow a |\,.\,|\overrightarrow b | = 4\,.\,|\overrightarrow c |$ ..... (iv)

(ii) $ \Rightarrow |\overrightarrow b |\,.\,|\overrightarrow c | = 9\,.\,|\overrightarrow a |$ ..... (v)

(iii) $ \Rightarrow |\overrightarrow c |\,.\,|\overrightarrow a | = \alpha \,.\,|\overrightarrow b |$ .... (vi)

Multiplying (iv), (v) and (vi)

$ \Rightarrow |\overrightarrow a |\,.\,|\overrightarrow b |\,.\,|\overrightarrow c | = 36\alpha $ ..... (vii)

Dividing (vii) by (iv) $ \Rightarrow |\overrightarrow c {|^2} = 9\alpha \Rightarrow |\overrightarrow c | = 3\sqrt \alpha $ ..... (viii)

Dividing (vii) by (v) $ \Rightarrow |\overrightarrow a {|^2} = 4\alpha \Rightarrow |\overrightarrow a | = 2\sqrt \alpha $

Dividing (viii) by (vi) $ \Rightarrow |\overrightarrow b {|^2} = 36 \Rightarrow |\overrightarrow b | = 6$

Now, as given, $3\sqrt \alpha + 2\sqrt \alpha + 6 = {1 \over {36}} \Rightarrow \sqrt \alpha = {{ - 43} \over {36}}$

$2{C_1} + {C_2} + 3{C_3} = 5$ ...... (i)

$3{C_1} + 3{C_2} + {C_3} = 0$ ...... (ii)

$\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right] = \left| {\matrix{ 2 & 1 & 3 \cr 3 & 3 & 1 \cr {{C_1}} & {{C_2}} & {{C_3}} \cr } } \right|$

$ = 2(3{C_3} - {C_2}) - 1(3{C_3} - {C_1}) + 3(3{C_2} - 3{C_1})$

$ = 3{C_3} + 7{C_2} - 8{C_1}$

$ \Rightarrow 8{C_1} - 7{C_2} - 3{C_3} = 0$ ...... (iii)

${C_1} = {{10} \over {122}},{C_2} = {{ - 85} \over {122}},{C_3} = {{225} \over {122}}$

So $122({C_1} + {C_2} + {C_3}) = 150$

Let $\overrightarrow a = x\widehat i = y\widehat j + z\widehat k$

So, $\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr x & y & z \cr 1 & 1 & \lambda \cr } } \right| = \widehat i(\lambda y - z) + \widehat j(z - \lambda x) + \widehat k(x - y)$

$ \Rightarrow \lambda y - z = 13,\,z - \lambda x = - 1,\,x - y = - 4$

and $x + y + \lambda z = - 21$

$\Rightarrow$ Clearly, $\lambda = 3$, $x = - 2$, $y = 2$ and $z = - 7$

So, $\overrightarrow b - \overrightarrow a = 3\widehat i - \widehat j + 10\widehat k$

and $\overrightarrow b + \overrightarrow a = - \widehat i + 3\widehat j - 4\widehat k$

$ \Rightarrow \left( {\overrightarrow b - \overrightarrow a } \right)\,.\,\left( {\widehat k - \widehat j} \right) + \left( {\overrightarrow b + \overrightarrow a } \right)\,.\,\left( {\widehat i - \widehat k} \right) = 11 + 3 = 14$

${\left| {\left( {\overrightarrow a - \overrightarrow b } \right) \times \left( {\overrightarrow a + \overrightarrow b } \right)} \right|^2} + 4{\left( {\overrightarrow a \,.\,\overrightarrow b } \right)^2}$

$ \Rightarrow {\left| {\overrightarrow a \times \overrightarrow a + \overrightarrow a \times \overrightarrow b - \overrightarrow b \times \overrightarrow a - \overrightarrow b \times \overrightarrow b } \right|^2} + 4{\left( {\overrightarrow a \,.\,\overrightarrow b } \right)^2}$

$ \Rightarrow {\left| {2\left( {\overrightarrow a \times \overrightarrow b } \right)} \right|^2} + 4{\left( {\overrightarrow a \,.\,\overrightarrow b } \right)^2}$

$ \Rightarrow 4{\left( {\overrightarrow a \times \overrightarrow b } \right)^2} + {\left( {\overrightarrow a \,.\,\overrightarrow b } \right)^2}$

$ \Rightarrow 4{\left| {\overrightarrow a } \right|^2}{\left| {\overrightarrow b } \right|^2} = 4\,.\,16\,.\,9 = 576$

Given, $\overrightarrow a = \widehat i + 2\widehat j - \widehat k$,

$\overrightarrow b = \widehat i - \widehat j$,

$\overrightarrow c = \widehat i - \widehat j - \widehat k$

$\overrightarrow r \times \overrightarrow a = \overrightarrow c \times \overrightarrow a $

$ \Rightarrow \overrightarrow r \times \overrightarrow a - \overrightarrow c \times \overrightarrow a = 0$

$ \Rightarrow (\overrightarrow r - \overrightarrow c ) \times \overrightarrow a = 0$

$\therefore$ $\overrightarrow r - \overrightarrow c = \lambda \overrightarrow a $

$ \Rightarrow \overrightarrow r = \lambda \overrightarrow a + \overrightarrow c $

$ \Rightarrow \overrightarrow r \,.\,\overrightarrow b = \lambda \overrightarrow a \,.\,\overrightarrow b + \overrightarrow c \,.\,\overrightarrow b $ (taking dot with $\overrightarrow b $)

$ \Rightarrow 0 = \lambda \overrightarrow a \,.\,\overrightarrow b + \overrightarrow c \,.\,\overrightarrow b $ [$\because$ $\overrightarrow r \,.\,\overrightarrow b = 0$]

$ \Rightarrow \lambda (\widehat i + 2\widehat j - \widehat k)\,.\,(\widehat i - \widehat j) + (\widehat i - \widehat j - \widehat k)\,.\,(\widehat i - \widehat j) = 0$

$ \Rightarrow \lambda (1 - 2) + 2 = 0$

$ \Rightarrow \lambda = 2$

$\therefore$ $\overrightarrow r = 2\overrightarrow a + \overrightarrow c $

$ \Rightarrow \overrightarrow r \,.\,\overrightarrow a = 2\overrightarrow a \,.\,\overrightarrow a + \overrightarrow c \,.\,\overrightarrow a $ [taking dot with ${\overrightarrow a }$]

$ = 2{\left| {\overrightarrow a } \right|^2} + \overrightarrow a \,.\,\overrightarrow c $

$ = 2(1 + 4 + 1) + (1 - 2 + 1)$

$ \Rightarrow \overrightarrow r \,.\,\overrightarrow a = 12$