Vector or Cross product of two vectors and its applications Questions in English

(C) Given $\vec{b} \cdot \vec{c} = |\vec{b}| |\vec{c}| \cos(\frac{\pi}{3}) = 10$.
Since $|\vec{b}| = 5$,we have $5 |\vec{c}| (\frac{1}{2}) = 10$,which implies $|\vec{c}| = 4$.
Now,the magnitude of the cross product is $|\vec{b} \times \vec{c}| = |\vec{b}| |\vec{c}| \sin(\frac{\pi}{3}) = 5 \times 4 \times \frac{\sqrt{3}}{2} = 10\sqrt{3}$.
Since $\vec{a}$ is perpendicular to $\vec{b} \times \vec{c}$,the angle between $\vec{a}$ and $\vec{b} \times \vec{c}$ is $\frac{\pi}{2}$.
Thus,$|\vec{a} \times (\vec{b} \times \vec{c})| = |\vec{a}| |\vec{b} \times \vec{c}| \sin(\frac{\pi}{2}) = \sqrt{3} \times 10\sqrt{3} \times 1 = 30$.

(A) Given vectors are $\vec{a} = 2\hat{i} + \hat{j} + 3\hat{k}$ and $\vec{b} = 3\hat{i} + 5\hat{j} - 2\hat{k}$.
The cross product $\vec{a} \times \vec{b}$ is calculated using the determinant:
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 3 \\ 3 & 5 & -2 \end{vmatrix}$
$= \hat{i}((1)(-2) - (3)(5)) - \hat{j}((2)(-2) - (3)(3)) + \hat{k}((2)(5) - (1)(3))$
$= \hat{i}(-2 - 15) - \hat{j}(-4 - 9) + \hat{k}(10 - 3)$
$= -17\hat{i} + 13\hat{j} + 7\hat{k}$
Now,the magnitude $|\vec{a} \times \vec{b}|$ is:
$|\vec{a} \times \vec{b}| = \sqrt{(-17)^2 + (13)^2 + (7)^2}$
$= \sqrt{289 + 169 + 49}$
$= \sqrt{507}$

(A) Given $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}$.
First,calculate $\vec{a} + \vec{b} = (1+1)\hat{i} + (1+2)\hat{j} + (1+3)\hat{k} = 2\hat{i} + 3\hat{j} + 4\hat{k}$.
Next,calculate $\vec{a} - \vec{b} = (1-1)\hat{i} + (1-2)\hat{j} + (1-3)\hat{k} = 0\hat{i} - \hat{j} - 2\hat{k}$.
$A$ vector perpendicular to both $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$ is their cross product $\vec{c} = (\vec{a} + \vec{b}) \times (\vec{a} - \vec{b})$.
$\vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 0 & -1 & -2 \end{vmatrix} = \hat{i}(-6 - (-4)) - \hat{j}(-4 - 0) + \hat{k}(-2 - 0) = -2\hat{i} + 4\hat{j} - 2\hat{k}$.
The magnitude is $|\vec{c}| = \sqrt{(-2)^2 + 4^2 + (-2)^2} = \sqrt{4 + 16 + 4} = \sqrt{24} = 2\sqrt{6}$.
The required unit vector is $\pm \frac{\vec{c}}{|\vec{c}|} = \pm \frac{-2\hat{i} + 4\hat{j} - 2\hat{k}}{2\sqrt{6}} = \pm (\frac{-1}{\sqrt{6}}\hat{i} + \frac{2}{\sqrt{6}}\hat{j} - \frac{1}{\sqrt{6}}\hat{k})$.
Comparing with the options,option $A$ is correct.

(B) The vertices of the triangle are $A(1, 1, 1)$,$B(1, 2, 3)$,and $C(2, 3, 1)$.
First,we find the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$:
$\overrightarrow{AB} = (1-1)\hat{i} + (2-1)\hat{j} + (3-1)\hat{k} = 0\hat{i} + 1\hat{j} + 2\hat{k} = \hat{j} + 2\hat{k}$
$\overrightarrow{AC} = (2-1)\hat{i} + (3-1)\hat{j} + (1-1)\hat{k} = 1\hat{i} + 2\hat{j} + 0\hat{k} = \hat{i} + 2\hat{j}$
The area of the triangle is given by $\frac{1}{2} |\overrightarrow{AB} \times \overrightarrow{AC}|$.
Now,calculate the cross product $\overrightarrow{AB} \times \overrightarrow{AC}$:
$\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 2 \\ 1 & 2 & 0 \end{vmatrix}$
$= \hat{i}(0 - 4) - \hat{j}(0 - 2) + \hat{k}(0 - 1) = -4\hat{i} + 2\hat{j} - \hat{k}$.
Next,find the magnitude of the cross product:
$|\overrightarrow{AB} \times \overrightarrow{AC}| = \sqrt{(-4)^2 + (2)^2 + (-1)^2} = \sqrt{16 + 4 + 1} = \sqrt{21}$.
Thus,the area of the triangle is $\frac{1}{2} \sqrt{21}$ square units.

(A) The area of a parallelogram with adjacent sides represented by vectors $\vec{a}$ and $\vec{b}$ is given by the magnitude of their cross product,i.e.,$|\vec{a} \times \vec{b}|$.
First,we calculate the cross product $\vec{a} \times \vec{b}$ using the determinant method:
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 4 \\ 1 & -1 & 1 \end{vmatrix}$
$= \hat{i}(1(1) - 4(-1)) - \hat{j}(3(1) - 4(1)) + \hat{k}(3(-1) - 1(1))$
$= \hat{i}(1 + 4) - \hat{j}(3 - 4) + \hat{k}(-3 - 1)$
$= 5\hat{i} + \hat{j} - 4\hat{k}$.
Now,we find the magnitude of the resulting vector:
$|\vec{a} \times \vec{b}| = \sqrt{(5)^2 + (1)^2 + (-4)^2}$
$= \sqrt{25 + 1 + 16} = \sqrt{42}$.
Thus,the area of the parallelogram is $\sqrt{42}$ square units.

(A) We have,$\vec{a} = \hat{i} - 7\hat{j} + 7\hat{k}$ and $\vec{b} = 3\hat{i} - 2\hat{j} + 2\hat{k}$.
The cross product $\vec{a} \times \vec{b}$ is given by the determinant:
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -7 & 7 \\ 3 & -2 & 2 \end{vmatrix}$
Expanding the determinant:
$= \hat{i}((-7)(2) - (7)(-2)) - \hat{j}((1)(2) - (7)(3)) + \hat{k}((1)(-2) - (-7)(3))$
$= \hat{i}(-14 + 14) - \hat{j}(2 - 21) + \hat{k}(-2 + 21)$
$= 0\hat{i} + 19\hat{j} + 19\hat{k}$
Now,find the magnitude $|\vec{a} \times \vec{b}| = \sqrt{0^2 + 19^2 + 19^2}$
$= \sqrt{19^2 + 19^2} = \sqrt{2 \times 19^2} = 19\sqrt{2}$.

(A) Given $\vec{a}=3 \hat{i}+2 \hat{j}+2 \hat{k}$ and $\vec{b}=\hat{i}+2 \hat{j}-2 \hat{k}$.
First,calculate the sum and difference of the vectors:
$\vec{a}+\vec{b} = (3+1)\hat{i} + (2+2)\hat{j} + (2-2)\hat{k} = 4\hat{i} + 4\hat{j}$
$\vec{a}-\vec{b} = (3-1)\hat{i} + (2-2)\hat{j} + (2-(-2))\hat{k} = 2\hat{i} + 4\hat{k}$
Now,find the cross product $(\vec{a}+\vec{b}) \times (\vec{a}-\vec{b})$:
$(\vec{a}+\vec{b}) \times (\vec{a}-\vec{b}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 4 & 0 \\ 2 & 0 & 4 \end{vmatrix}$
$= \hat{i}(16-0) - \hat{j}(16-0) + \hat{k}(0-8) = 16\hat{i} - 16\hat{j} - 8\hat{k}$
Calculate the magnitude of the cross product:
$|(\vec{a}+\vec{b}) \times (\vec{a}-\vec{b})| = \sqrt{16^2 + (-16)^2 + (-8)^2} = \sqrt{256 + 256 + 64} = \sqrt{576} = 24$
The unit vector perpendicular to both is given by:
$\pm \frac{(\vec{a}+\vec{b}) \times (\vec{a}-\vec{b})}{|(\vec{a}+\vec{b}) \times (\vec{a}-\vec{b})|} = \pm \frac{16\hat{i} - 16\hat{j} - 8\hat{k}}{24}$
$= \pm \frac{2\hat{i} - 2\hat{j} - \hat{k}}{3} = \pm \frac{2}{3}\hat{i} \mp \frac{2}{3}\hat{j} \mp \frac{1}{3}\hat{k}$

(N/A) We use the distributive property of the vector cross product over addition:
$(\vec{a}-\vec{b}) \times(\vec{a}+\vec{b}) = (\vec{a}-\vec{b}) \times \vec{a} + (\vec{a}-\vec{b}) \times \vec{b}$
$= (\vec{a} \times \vec{a}) - (\vec{b} \times \vec{a}) + (\vec{a} \times \vec{b}) - (\vec{b} \times \vec{b})$
Since the cross product of any vector with itself is the zero vector,i.e.,$\vec{a} \times \vec{a} = \vec{0}$ and $\vec{b} \times \vec{b} = \vec{0}$,and using the anti-commutative property $\vec{b} \times \vec{a} = -(\vec{a} \times \vec{b})$:
$= \vec{0} - (-\vec{a} \times \vec{b}) + (\vec{a} \times \vec{b}) - \vec{0}$
$= (\vec{a} \times \vec{b}) + (\vec{a} \times \vec{b})$
$= 2(\vec{a} \times \vec{b})$
Hence,the given expression is proved.

(A) Given the cross product of two vectors is the zero vector: $(2 \hat{i}+6 \hat{j}+27 \hat{k}) \times(\hat{i}+\lambda \hat{j}+\mu \hat{k})=\overrightarrow{0}$.
We express the cross product as a determinant:
$\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 6 & 27 \\ 1 & \lambda & \mu\end{array}\right|=0 \hat{i}+0 \hat{j}+0 \hat{k}$.
Expanding the determinant:
$\hat{i}(6 \mu-27 \lambda)-\hat{j}(2 \mu-27)+\hat{k}(2 \lambda-6)=0 \hat{i}+0 \hat{j}+0 \hat{k}$.
Comparing the components on both sides:
$1) \ 6 \mu-27 \lambda=0$
$2) \ -(2 \mu-27)=0 \Rightarrow 2 \mu=27 \Rightarrow \mu=\frac{27}{2}$
$3) \ 2 \lambda-6=0 \Rightarrow 2 \lambda=6 \Rightarrow \lambda=3$.
Substituting $\lambda=3$ into the first equation: $6(\frac{27}{2}) - 27(3) = 81 - 81 = 0$,which is consistent.
Thus,the values are $\lambda=3$ and $\mu=\frac{27}{2}$.

(N/A) Given $\vec{a}=a_{1} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k}$,$\vec{b}=b_{1} \hat{i}+b_{2} \hat{j}+b_{3} \hat{k}$,and $\vec{c}=c_{1} \hat{i}+c_{2} \hat{j}+c_{3} \hat{k}$.
First,calculate $(\vec{b}+\vec{c}) = (b_{1}+c_{1}) \hat{i} + (b_{2}+c_{2}) \hat{j} + (b_{3}+c_{3}) \hat{k}$.
Now,$\vec{a} \times(\vec{b}+\vec{c}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ b_{1}+c_{1} & b_{2}+c_{2} & b_{3}+c_{3} \end{vmatrix}$
$= \hat{i}[a_{2}(b_{3}+c_{3}) - a_{3}(b_{2}+c_{2})] - \hat{j}[a_{1}(b_{3}+c_{3}) - a_{3}(b_{1}+c_{1})] + \hat{k}[a_{1}(b_{2}+c_{2}) - a_{2}(b_{1}+c_{1})]$
$= \hat{i}[a_{2}b_{3} + a_{2}c_{3} - a_{3}b_{2} - a_{3}c_{2}] + \hat{j}[a_{3}b_{1} + a_{3}c_{1} - a_{1}b_{3} - a_{1}c_{3}] + \hat{k}[a_{1}b_{2} + a_{1}c_{2} - a_{2}b_{1} - a_{2}c_{1}] \dots (1)$
Next,$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \end{vmatrix} = \hat{i}(a_{2}b_{3}-a_{3}b_{2}) + \hat{j}(a_{3}b_{1}-a_{1}b_{3}) + \hat{k}(a_{1}b_{2}-a_{2}b_{1}) \dots (2)$
And $\vec{a} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ c_{1} & c_{2} & c_{3} \end{vmatrix} = \hat{i}(a_{2}c_{3}-a_{3}c_{2}) + \hat{j}(a_{3}c_{1}-a_{1}c_{3}) + \hat{k}(a_{1}c_{2}-a_{2}c_{1}) \dots (3)$
Adding $(2)$ and $(3)$ gives:
$(\vec{a} \times \vec{b}) + (\vec{a} \times \vec{c}) = \hat{i}(a_{2}b_{3} + a_{2}c_{3} - a_{3}b_{2} - a_{3}c_{2}) + \hat{j}(a_{3}b_{1} + a_{3}c_{1} - a_{1}b_{3} - a_{1}c_{3}) + \hat{k}(a_{1}b_{2} + a_{1}c_{2} - a_{2}b_{1} - a_{2}c_{1}) \dots (4)$
Comparing $(1)$ and $(4)$,we see that $\vec{a} \times(\vec{b}+\vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c}$. Hence proved.

(B) The converse of the statement is: If $\vec{a} \times \vec{b}=\overrightarrow{0},$ then either $\vec{a}=\overrightarrow{0}$ or $\vec{b}=\overrightarrow{0}.$
This converse is not necessarily true.
Consider two non-zero parallel vectors $\vec{a}$ and $\vec{b}.$
Let $\vec{a}=2 \hat{i}+3 \hat{j}+4 \hat{k}$ and $\vec{b}=4 \hat{i}+6 \hat{j}+8 \hat{k}.$
Calculating the cross product:
$\vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 4 & 6 & 8\end{array}\right| = \hat{i}(24-24) - \hat{j}(16-16) + \hat{k}(12-12) = \overrightarrow{0}.$
Here,$|\vec{a}| = \sqrt{2^2+3^2+4^2} = \sqrt{29} \neq 0$ and $|\vec{b}| = \sqrt{4^2+6^2+8^2} = \sqrt{116} \neq 0.$
Since $\vec{a} \times \vec{b} = \overrightarrow{0}$ even though $\vec{a} \neq \overrightarrow{0}$ and $\vec{b} \neq \overrightarrow{0},$ the converse is not true.

(A) The vertices of triangle $ABC$ are given as $A(1,1,2), B(2,3,5)$ and $C(1,5,5)$.
The vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are:
$\overrightarrow{AB} = (2-1)\hat{i} + (3-1)\hat{j} + (5-2)\hat{k} = \hat{i} + 2\hat{j} + 3\hat{k}$
$\overrightarrow{AC} = (1-1)\hat{i} + (5-1)\hat{j} + (5-2)\hat{k} = 0\hat{i} + 4\hat{j} + 3\hat{k}$
The area of $\triangle ABC$ is given by $\frac{1}{2} |\overrightarrow{AB} \times \overrightarrow{AC}|$.
$\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 0 & 4 & 3 \end{vmatrix} = \hat{i}(6-12) - \hat{j}(3-0) + \hat{k}(4-0) = -6\hat{i} - 3\hat{j} + 4\hat{k}$.
$|\overrightarrow{AB} \times \overrightarrow{AC}| = \sqrt{(-6)^2 + (-3)^2 + 4^2} = \sqrt{36 + 9 + 16} = \sqrt{61}$.
Therefore,the area of $\triangle ABC = \frac{\sqrt{61}}{2}$ square units.

(A) The area of a parallelogram with adjacent sides represented by vectors $\vec{a}$ and $\vec{b}$ is given by the magnitude of their cross product,$|\vec{a} \times \vec{b}|$.
Given vectors are $\vec{a} = \hat{i} - \hat{j} + 3\hat{k}$ and $\vec{b} = 2\hat{i} - 7\hat{j} + \hat{k}$.
First,calculate the cross product $\vec{a} \times \vec{b}$:
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 3 \\ 2 & -7 & 1 \end{vmatrix}$
$= \hat{i}((-1)(1) - (3)(-7)) - \hat{j}((1)(1) - (3)(2)) + \hat{k}((1)(-7) - (-1)(2))$
$= \hat{i}(-1 + 21) - \hat{j}(1 - 6) + \hat{k}(-7 + 2)$
$= 20\hat{i} + 5\hat{j} - 5\hat{k}$.
Now,calculate the magnitude $|\vec{a} \times \vec{b}|$:
$|\vec{a} \times \vec{b}| = \sqrt{(20)^2 + (5)^2 + (-5)^2}$
$= \sqrt{400 + 25 + 25}$
$= \sqrt{450}$
$= \sqrt{225 \times 2}$
$= 15\sqrt{2}$.
Thus,the area of the parallelogram is $15\sqrt{2}$ square units.

(B) The position vectors of vertices $A, B, C,$ and $D$ of rectangle $ABCD$ are given as:
$\overrightarrow{OA} = -\hat{i} + \frac{1}{2}\hat{j} + 4\hat{k}, \overrightarrow{OB} = \hat{i} + \frac{1}{2}\hat{j} + 4\hat{k}$
$\overrightarrow{OC} = \hat{i} - \frac{1}{2}\hat{j} + 4\hat{k}, \overrightarrow{OD} = -\hat{i} - \frac{1}{2}\hat{j} + 4\hat{k}$
The adjacent sides $\overrightarrow{AB}$ and $\overrightarrow{BC}$ are calculated as:
$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (1 - (-1))\hat{i} + (\frac{1}{2} - \frac{1}{2})\hat{j} + (4 - 4)\hat{k} = 2\hat{i}$
$\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = (1 - 1)\hat{i} + (-\frac{1}{2} - \frac{1}{2})\hat{j} + (4 - 4)\hat{k} = -\hat{j}$
The area of the rectangle is given by the magnitude of the cross product of its adjacent sides:
$\text{Area} = |\overrightarrow{AB} \times \overrightarrow{BC}|$
$\overrightarrow{AB} \times \overrightarrow{BC} = (2\hat{i}) \times (-\hat{j}) = -2(\hat{i} \times \hat{j}) = -2\hat{k}$
$|\overrightarrow{AB} \times \overrightarrow{BC}| = |-2\hat{k}| = 2$
Thus,the area of the rectangle is $2$ square units.

(A) Let the adjacent sides be $\vec{a} = 2 \hat{i}-4 \hat{j}+5 \hat{k}$ and $\vec{b} = \hat{i}-2 \hat{j}-3 \hat{k}$.
The diagonal of the parallelogram is given by $\vec{d} = \vec{a} + \vec{b} = (2+1) \hat{i} + (-4-2) \hat{j} + (5-3) \hat{k} = 3 \hat{i} - 6 \hat{j} + 2 \hat{k}$.
The magnitude of the diagonal is $|\vec{d}| = \sqrt{3^2 + (-6)^2 + 2^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7$.
The unit vector parallel to the diagonal is $\frac{\vec{d}}{|\vec{d}|} = \frac{3 \hat{i} - 6 \hat{j} + 2 \hat{k}}{7} = \frac{3}{7} \hat{i} - \frac{6}{7} \hat{j} + \frac{2}{7} \hat{k}$.
The area of the parallelogram is given by $|\vec{a} \times \vec{b}|$.
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -4 & 5 \\ 1 & -2 & -3 \end{vmatrix} = \hat{i}(12 - (-10)) - \hat{j}(-6 - 5) + \hat{k}(-4 - (-4)) = 22 \hat{i} + 11 \hat{j} + 0 \hat{k}$.
Area $= |22 \hat{i} + 11 \hat{j}| = \sqrt{22^2 + 11^2} = \sqrt{484 + 121} = \sqrt{605} = 11 \sqrt{5}$ square units.

(A) Let $\vec{d}=d_{1} \hat{i}+d_{2} \hat{j}+d_{3} \hat{k}$.
Since $\vec{d}$ is perpendicular to both $\vec{a}$ and $\vec{b}$,we have $\vec{d} \cdot \vec{a} = 0$ and $\vec{d} \cdot \vec{b} = 0$.
$\vec{d} \cdot \vec{a} = d_{1} + 4d_{2} + 2d_{3} = 0$ .....$(i)$
$\vec{d} \cdot \vec{b} = 3d_{1} - 2d_{2} + 7d_{3} = 0$ .....$(ii)$
Also,it is given that $\vec{c} \cdot \vec{d} = 15$,so $2d_{1} - d_{2} + 4d_{3} = 15$ .....$(iii)$
From $(i)$,$d_{1} = -4d_{2} - 2d_{3}$. Substituting this into $(ii)$:
$3(-4d_{2} - 2d_{3}) - 2d_{2} + 7d_{3} = 0 \Rightarrow -12d_{2} - 6d_{3} - 2d_{2} + 7d_{3} = 0 \Rightarrow -14d_{2} + d_{3} = 0 \Rightarrow d_{3} = 14d_{2}$.
Substituting $d_{3} = 14d_{2}$ into $(i)$:
$d_{1} + 4d_{2} + 2(14d_{2}) = 0 \Rightarrow d_{1} + 4d_{2} + 28d_{2} = 0 \Rightarrow d_{1} = -32d_{2}$.
Now substitute $d_{1}$ and $d_{3}$ into $(iii)$:
$2(-32d_{2}) - d_{2} + 4(14d_{2}) = 15 \Rightarrow -64d_{2} - d_{2} + 56d_{2} = 15 \Rightarrow -9d_{2} = 15 \Rightarrow d_{2} = -\frac{15}{9} = -\frac{5}{3}$.
Then $d_{1} = -32(-\frac{5}{3}) = \frac{160}{3}$ and $d_{3} = 14(-\frac{5}{3}) = -\frac{70}{3}$.
Thus,$\vec{d} = \frac{160}{3} \hat{i} - \frac{5}{3} \hat{j} - \frac{70}{3} \hat{k} = \frac{1}{3}(160 \hat{i} - 5 \hat{j} - 70 \hat{k})$.

(A) Let $\vec{a} = \hat{i} + 2 \hat{j} + \hat{k}$ and $\vec{b} = -\hat{i} + 3 \hat{j} + 4 \hat{k}$.
The vector perpendicular to the plane containing $\vec{a}$ and $\vec{b}$ is given by the cross product $\vec{a} \times \vec{b}$.
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 1 \\ -1 & 3 & 4 \end{vmatrix} = \hat{i}(8 - 3) - \hat{j}(4 + 1) + \hat{k}(3 + 2) = 5\hat{i} - 5\hat{j} + 5\hat{k}$.
The magnitude of this vector is $|\vec{a} \times \vec{b}| = \sqrt{5^2 + (-5)^2 + 5^2} = \sqrt{25 + 25 + 25} = \sqrt{75} = 5\sqrt{3}$.
The unit vector perpendicular to the plane is $\hat{n} = \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} = \frac{5\hat{i} - 5\hat{j} + 5\hat{k}}{5\sqrt{3}} = \frac{1}{\sqrt{3}}(\hat{i} - \hat{j} + \hat{k})$.
Vectors of magnitude $10\sqrt{3}$ perpendicular to the plane are given by $\pm 10\sqrt{3} \times \hat{n} = \pm 10\sqrt{3} \times \frac{1}{\sqrt{3}}(\hat{i} - \hat{j} + \hat{k}) = \pm 10(\hat{i} - \hat{j} + \hat{k})$.
Thus,the required vectors are $10\hat{i} - 10\hat{j} + 10\hat{k}$ and $-10\hat{i} + 10\hat{j} - 10\hat{k}$.

(A) Let the three sides of the triangle $BC, CA$,and $AB$ be represented by vectors $\vec{a}, \vec{b}$,and $\vec{c}$,respectively (as shown in the figure).
We have $\vec{a} + \vec{b} + \vec{c} = \vec{0}$.
This implies $\vec{a} + \vec{b} = -\vec{c}$.
Taking the cross product with $\vec{a}$ on both sides: $\vec{a} \times (\vec{a} + \vec{b}) = \vec{a} \times (-\vec{c}) \Rightarrow \vec{a} \times \vec{a} + \vec{a} \times \vec{b} = -\vec{a} \times \vec{c} \Rightarrow \vec{0} + \vec{a} \times \vec{b} = \vec{c} \times \vec{a} \Rightarrow \vec{a} \times \vec{b} = \vec{c} \times \vec{a}$.
Similarly,taking the cross product with $\vec{b}$ on both sides of $\vec{a} + \vec{b} = -\vec{c}$: $(\vec{a} + \vec{b}) \times \vec{b} = -\vec{c} \times \vec{b} \Rightarrow \vec{a} \times \vec{b} + \vec{b} \times \vec{b} = \vec{b} \times \vec{c} \Rightarrow \vec{a} \times \vec{b} = \vec{b} \times \vec{c}$.
Thus,$\vec{a} \times \vec{b} = \vec{b} \times \vec{c} = \vec{c} \times \vec{a}$.
Taking the magnitude of these vectors: $|\vec{a} \times \vec{b}| = |\vec{b} \times \vec{c}| = |\vec{c} \times \vec{a}|$.
Using the definition of the cross product $|\vec{u} \times \vec{v}| = |\vec{u}||\vec{v}| \sin \theta$,where $\theta$ is the angle between the vectors:
$|\vec{a}||\vec{b}| \sin(\pi - C) = |\vec{b}||\vec{c}| \sin(\pi - A) = |\vec{c}||\vec{a}| \sin(\pi - B)$.
Since $\sin(\pi - \theta) = \sin \theta$,we get:
$ab \sin C = bc \sin A = ca \sin B$.
Dividing the entire equation by $abc$,we get:
$\frac{ab \sin C}{abc} = \frac{bc \sin A}{abc} = \frac{ca \sin B}{abc} \Rightarrow \frac{\sin C}{c} = \frac{\sin A}{a} = \frac{\sin B}{b}$.
Hence,$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$.

(A) Let $\vec{a} = 2 \hat{i}-\hat{j}+2 \hat{k}$ and $\vec{b} = 4 \hat{i}-\hat{j}+3 \hat{k}$.
$A$ vector perpendicular to both $\vec{a}$ and $\vec{b}$ is given by the cross product $\vec{r} = \vec{a} \times \vec{b}$.
$\vec{r} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 2 \\ 4 & -1 & 3 \end{vmatrix}$
$= \hat{i}(-3 - (-2)) - \hat{j}(6 - 8) + \hat{k}(-2 - (-4))$
$= \hat{i}(-1) - \hat{j}(-2) + \hat{k}(2) = -\hat{i} + 2 \hat{j} + 2 \hat{k}$.
The magnitude of $\vec{r}$ is $|\vec{r}| = \sqrt{(-1)^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.
$A$ unit vector in the direction of $\vec{r}$ is $\hat{r} = \frac{\vec{r}}{|\vec{r}|} = \frac{-\hat{i} + 2 \hat{j} + 2 \hat{k}}{3}$.
$A$ vector of magnitude $6$ in the direction of $\vec{r}$ is $6 \hat{r} = 6 \left( \frac{-\hat{i} + 2 \hat{j} + 2 \hat{k}}{3} \right) = 2(-\hat{i} + 2 \hat{j} + 2 \hat{k}) = -2 \hat{i} + 4 \hat{j} + 4 \hat{k}$.
Note: The vector $-6 \hat{r} = 2 \hat{i} - 4 \hat{j} - 4 \hat{k}$ is also a valid answer.

(N/A) Given,$\vec{a}+\vec{b}+\vec{c}=0$
$\Rightarrow \vec{b}=-\vec{c}-\vec{a}$
Now,$\vec{a} \times \vec{b}=\vec{a} \times(-\vec{c}-\vec{a})$
$=(\vec{a} \times(-\vec{c}))+(\vec{a} \times(-\vec{a}))$
$=-( \vec{a} \times \vec{c} ) - 0 = \vec{c} \times \vec{a} \ldots (i)$
Also,$\vec{b} \times \vec{c}=(-\vec{c}-\vec{a}) \times \vec{c}$
$=(-\vec{c} \times \vec{c})+(-\vec{a} \times \vec{c})$
$=0 - (\vec{a} \times \vec{c}) = \vec{c} \times \vec{a} \ldots (ii)$
From equations $(i)$ and $(ii)$,we get $\vec{a} \times \vec{b}=\vec{b} \times \vec{c}=\vec{c} \times \vec{a}$.
Geometrical Interpretation:
If $\vec{a}, \vec{b}, \vec{c}$ are the sides of a triangle $ABC$ taken in order,then $\vec{a}+\vec{b}+\vec{c}=0$. The magnitude of the cross product of two vectors represents the area of the parallelogram formed by them. Since $\vec{a} \times \vec{b} = \vec{b} \times \vec{c} = \vec{c} \times \vec{a}$,the areas of the parallelograms formed by any two of these vectors as adjacent sides are equal. This is consistent with the fact that these vectors form a triangle,and the cross products represent twice the area of the triangle formed by the vectors.

(N/A) The position vectors of the vertices are $\vec{A} = \hat{i} + 2\hat{j} + 3\hat{k}$,$\vec{B} = 2\hat{i} - \hat{j} + 4\hat{k}$,and $\vec{C} = 4\hat{i} + 5\hat{j} - \hat{k}$.
First,we find the vectors $\vec{AB}$ and $\vec{AC}$:
$\vec{AB} = \vec{B} - \vec{A} = (2-1)\hat{i} + (-1-2)\hat{j} + (4-3)\hat{k} = \hat{i} - 3\hat{j} + \hat{k}$
$\vec{AC} = \vec{C} - \vec{A} = (4-1)\hat{i} + (5-2)\hat{j} + (-1-3)\hat{k} = 3\hat{i} + 3\hat{j} - 4\hat{k}$
Now,calculate the cross product $\vec{AB} \times \vec{AC}$:
$\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 1 \\ 3 & 3 & -4 \end{vmatrix}$
$= \hat{i}((-3)(-4) - (1)(3)) - \hat{j}((1)(-4) - (1)(3)) + \hat{k}((1)(3) - (-3)(3))$
$= \hat{i}(12 - 3) - \hat{j}(-4 - 3) + \hat{k}(3 + 9)$
$= 9\hat{i} + 7\hat{j} + 12\hat{k}$
Next,find the magnitude of the cross product:
$|\vec{AB} \times \vec{AC}| = \sqrt{9^2 + 7^2 + 12^2} = \sqrt{81 + 49 + 144} = \sqrt{274}$
The area of triangle $ABC$ is given by $\frac{1}{2} |\vec{AB} \times \vec{AC}|$:
Area $= \frac{1}{2} \sqrt{274} \text{ sq units}$.

(N/A) Let $ABCD$ and $ABFE$ be two parallelograms on the same base $AB$ and between the same parallel lines $AB$ and $DF$.
Let $\overrightarrow{AB} = \vec{a}$ and $\overrightarrow{AD} = \vec{b}$.
The area of parallelogram $ABCD$ is given by the magnitude of the cross product of its adjacent sides:
$\text{Area}(ABCD) = |\overrightarrow{AB} \times \overrightarrow{AD}| = |\vec{a} \times \vec{b}|$.
Now,for parallelogram $ABFE$,the adjacent sides are $\overrightarrow{AB}$ and $\overrightarrow{AE}$.
Since $D, E, C, F$ lie on the same line parallel to $AB$,we can write $\overrightarrow{AE} = \overrightarrow{AD} + \overrightarrow{DE}$.
Since $\overrightarrow{DE}$ is parallel to $\overrightarrow{AB}$,we have $\overrightarrow{DE} = k\vec{a}$ for some scalar $k$.
Thus,$\overrightarrow{AE} = \vec{b} + k\vec{a}$.
The area of parallelogram $ABFE$ is:
$\text{Area}(ABFE) = |\overrightarrow{AB} \times \overrightarrow{AE}|$
$= |\vec{a} \times (\vec{b} + k\vec{a})|$
$= |(\vec{a} \times \vec{b}) + (\vec{a} \times k\vec{a})|$
$= |(\vec{a} \times \vec{b}) + k(\vec{a} \times \vec{a})|$.
Since the cross product of any vector with itself is zero $(\vec{a} \times \vec{a} = \vec{0})$:
$\text{Area}(ABFE) = |\vec{a} \times \vec{b} + \vec{0}| = |\vec{a} \times \vec{b}|$.
Therefore,$\text{Area}(ABFE) = \text{Area}(ABCD)$.
Hence,it is proved that parallelograms on the same base and between the same parallels are equal in area.

(B) Given points $P(3, -1, 2)$ and $Q(1, 2, -4)$.
Direction vectors of lines $PR$ and $QS$ are $\vec{v_1} = 4\hat{i} - \hat{j} + 2\hat{k}$ and $\vec{v_2} = -2\hat{i} + \hat{j} - 2\hat{k}$.
The normal vector $\vec{n}$ to the plane containing $P, T, Q$ is $\vec{v_1} \times \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -1 & 2 \\ -2 & 1 & -2 \end{vmatrix} = \hat{i}(2-2) - \hat{j}(-8+4) + \hat{k}(4-2) = 0\hat{i} + 4\hat{j} + 2\hat{k}$.
Unit normal vector $\hat{n} = \frac{4\hat{j} + 2\hat{k}}{\sqrt{0^2 + 4^2 + 2^2}} = \frac{4\hat{j} + 2\hat{k}}{\sqrt{20}} = \frac{2\hat{j} + \hat{k}}{\sqrt{5}}$.
For intersection point $T$,line $PT: \vec{r} = (3\hat{i} - \hat{j} + 2\hat{k}) + \lambda(4\hat{i} - \hat{j} + 2\hat{k})$ and line $QT: \vec{r} = (\hat{i} + 2\hat{j} - 4\hat{k}) + \mu(-2\hat{i} + \hat{j} - 2\hat{k})$.
Equating coordinates: $3+4\lambda = 1-2\mu \Rightarrow 4\lambda + 2\mu = -2 \Rightarrow 2\lambda + \mu = -1$.
$-1-\lambda = 2+\mu \Rightarrow \lambda + \mu = -3$.
Subtracting equations: $\lambda = 2$,then $\mu = -5$.
Point $T = (3+4(2), -1-(2), 2+2(2)) = (11, -3, 6)$.
Vector $\vec{OA} = \vec{OT} \pm |\vec{TA}|\hat{n} = (11\hat{i} - 3\hat{j} + 6\hat{k}) \pm \sqrt{5} \left(\frac{2\hat{j} + \hat{k}}{\sqrt{5}}\right) = (11\hat{i} - 3\hat{j} + 6\hat{k}) \pm (2\hat{j} + \hat{k})$.
Case $1$: $\vec{OA} = 11\hat{i} - \hat{j} + 7\hat{k} \Rightarrow |\vec{OA}| = \sqrt{121 + 1 + 49} = \sqrt{171}$.
Case $2$: $\vec{OA} = 11\hat{i} - 5\hat{j} + 5\hat{k} \Rightarrow |\vec{OA}| = \sqrt{121 + 25 + 25} = \sqrt{171}$.

(N/A) Let the vertices of the triangle be $A, B, C$ with position vectors $\vec{a}, \vec{b}, \vec{c}$ respectively.
The vector area of $\Delta ABC$ is given by $\frac{1}{2}(\overrightarrow{AB} \times \overrightarrow{AC})$.
We have $\overrightarrow{AB} = \vec{b} - \vec{a}$ and $\overrightarrow{AC} = \vec{c} - \vec{a}$.
Vector area $= \frac{1}{2}[(\vec{b} - \vec{a}) \times (\vec{c} - \vec{a})]$
$= \frac{1}{2}[\vec{b} \times \vec{c} - \vec{b} \times \vec{a} - \vec{a} \times \vec{c} + \vec{a} \times \vec{a}]$
Since $\vec{a} \times \vec{a} = 0$,$-\vec{b} \times \vec{a} = \vec{a} \times \vec{b}$,and $-\vec{a} \times \vec{c} = \vec{c} \times \vec{a}$,we get:
Vector area $= \frac{1}{2}[\vec{b} \times \vec{c} + \vec{a} \times \vec{b} + \vec{c} \times \vec{a}]$.
For the points to be collinear,the area of the triangle must be zero.
Therefore,$\frac{1}{2}[\vec{b} \times \vec{c} + \vec{c} \times \vec{a} + \vec{a} \times \vec{b}] = 0$,which implies $\vec{b} \times \vec{c} + \vec{c} \times \vec{a} + \vec{a} \times \vec{b} = 0$.
The unit vector $\hat{n}$ normal to the plane is given by $\hat{n} = \frac{\overrightarrow{AB} \times \overrightarrow{AC}}{|\overrightarrow{AB} \times \overrightarrow{AC}|} = \frac{\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}}{|\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}|}$.

(N/A) Let $\vec{c}=x \hat{i}+y \hat{j}+z \hat{k}.$
Given $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}=\hat{j}-\hat{k}.$
For $\vec{a} \times \vec{c}=\vec{b},$
$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ x & y & z \end{vmatrix} = 0\hat{i} + 1\hat{j} - 1\hat{k}.$
Expanding the determinant,we get $\hat{i}(z-y) - \hat{j}(z-x) + \hat{k}(y-x) = 0\hat{i} + 1\hat{j} - 1\hat{k}.$
Comparing coefficients,we have:
$z-y=0 \implies z=y \quad (i)$
$x-z=1 \implies x=z+1 \quad (ii)$
$y-x=-1 \quad (iii)$
Also,$\vec{a} \cdot \vec{c}=3 \implies x+y+z=3 \quad (iv).$
Substituting $x=z+1$ and $y=z$ into $(iv)$:
$(z+1) + z + z = 3 \implies 3z+1=3 \implies 3z=2 \implies z=\frac{2}{3}.$
Thus,$y=\frac{2}{3}$ and $x=\frac{2}{3}+1=\frac{5}{3}.$
Therefore,$\vec{c}=\frac{5}{3}\hat{i}+\frac{2}{3}\hat{j}+\frac{2}{3}\hat{k} = \frac{1}{3}(5\hat{i}+2\hat{j}+2\hat{k}).$

(B) Since $\overrightarrow{c}$ is perpendicular to both $\overrightarrow{a}$ and $\overrightarrow{b},$ it must be parallel to their cross product $\overrightarrow{a}\times\overrightarrow{b}.$ Thus,we can write $\overrightarrow{c}=\lambda(\overrightarrow{a}\times\overrightarrow{b})$ for some scalar $\lambda.$
First,calculate $\overrightarrow{a}\times\overrightarrow{b}$:
$\overrightarrow{a}\times\overrightarrow{b}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -1 \\ 1 & 2 & 1 \end{vmatrix} = \hat{i}(1 - (-2)) - \hat{j}(1 - (-1)) + \hat{k}(2 - 1) = 3\hat{i}-2\hat{j}+\hat{k}.$
Given $\overrightarrow{c}\cdot(\hat{i}+\hat{j}+3\hat{k})=8,$ substitute $\overrightarrow{c}=\lambda(3\hat{i}-2\hat{j}+\hat{k})$:
$\lambda(3\hat{i}-2\hat{j}+\hat{k})\cdot(\hat{i}+\hat{j}+3\hat{k})=8$
$\lambda(3(1) + (-2)(1) + (1)(3)) = 8$
$\lambda(3 - 2 + 3) = 8 \Rightarrow 4\lambda = 8 \Rightarrow \lambda = 2.$
Now,calculate $\overrightarrow{c}\cdot(\overrightarrow{a}\times\overrightarrow{b})$:
$\overrightarrow{c}\cdot(\overrightarrow{a}\times\overrightarrow{b}) = \lambda(\overrightarrow{a}\times\overrightarrow{b})\cdot(\overrightarrow{a}\times\overrightarrow{b}) = \lambda|\overrightarrow{a}\times\overrightarrow{b}|^2.$
$|\overrightarrow{a}\times\overrightarrow{b}|^2 = 3^2 + (-2)^2 + 1^2 = 9 + 4 + 1 = 14.$
Therefore,$\overrightarrow{c}\cdot(\overrightarrow{a}\times\overrightarrow{b}) = 2 \times 14 = 28.$

(B) Given that $|\vec{a}|=|\vec{b}|$ and $\vec{a} \perp \vec{b}$.
Also,$|\vec{a} \times \vec{b}| = |\vec{a}|$.
Since $\vec{a} \perp \vec{b}$,we have $|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}| \sin 90^{\circ} = |\vec{a}||\vec{b}|$.
Substituting this into the given equation: $|\vec{a}||\vec{b}| = |\vec{a}|$.
Since $\vec{a}$ is a non-zero vector,$|\vec{a}| \neq 0$,so $|\vec{b}| = 1$. Consequently,$|\vec{a}| = 1$.
Thus,$\vec{a}$ and $\vec{b}$ are mutually perpendicular unit vectors.
Let $\vec{a} = \hat{i}$ and $\vec{b} = \hat{j}$. Then $\vec{a} \times \vec{b} = \hat{i} \times \hat{j} = \hat{k}$.
Let $\vec{v} = \vec{a} + \vec{b} + (\vec{a} \times \vec{b}) = \hat{i} + \hat{j} + \hat{k}$.
The angle $\theta$ between $\vec{v}$ and $\vec{a}$ is given by $\cos \theta = \frac{\vec{v} \cdot \vec{a}}{|\vec{v}| |\vec{a}|}$.
$\vec{v} \cdot \vec{a} = (\hat{i} + \hat{j} + \hat{k}) \cdot \hat{i} = 1$.
$|\vec{v}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$ and $|\vec{a}| = 1$.
So,$\cos \theta = \frac{1}{\sqrt{3} \times 1} = \frac{1}{\sqrt{3}}$.
Therefore,$\theta = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right)$.

(B) Given vectors are $\vec{a} = \hat{i} + \alpha \hat{j} + 3 \hat{k}$ and $\vec{b} = 3 \hat{i} - \alpha \hat{j} + \hat{k}$.
The area of the parallelogram with adjacent sides $\vec{a}$ and $\vec{b}$ is given by $|\vec{a} \times \vec{b}| = 8 \sqrt{3}$.
First,calculate the cross product $\vec{a} \times \vec{b}$:
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & \alpha & 3 \\ 3 & -\alpha & 1 \end{vmatrix} = \hat{i}(\alpha - (-3\alpha)) - \hat{j}(1 - 9) + \hat{k}(-\alpha - 3\alpha) = 4\alpha \hat{i} + 8 \hat{j} - 4\alpha \hat{k}$.
Now,find the magnitude:
$|\vec{a} \times \vec{b}| = \sqrt{(4\alpha)^2 + 8^2 + (-4\alpha)^2} = \sqrt{16\alpha^2 + 64 + 16\alpha^2} = \sqrt{32\alpha^2 + 64}$.
Equating to the given area:
$\sqrt{32\alpha^2 + 64} = 8 \sqrt{3} \Rightarrow 32\alpha^2 + 64 = 64 \times 3 = 192$.
$32\alpha^2 = 192 - 64 = 128 \Rightarrow \alpha^2 = 4$.
Finally,calculate the dot product $\vec{a} \cdot \vec{b}$:
$\vec{a} \cdot \vec{b} = (1)(3) + (\alpha)(-\alpha) + (3)(1) = 3 - \alpha^2 + 3 = 6 - \alpha^2$.
Substituting $\alpha^2 = 4$,we get $\vec{a} \cdot \vec{b} = 6 - 4 = 2$.

(D) Given $\vec{a} \cdot \vec{b} = 0$,we have $(1)(1) + (5)(3) + (\alpha)(\beta) = 0$,which implies $1 + 15 + \alpha\beta = 0$,so $\alpha\beta = -16$.
Next,calculate $\vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 3 & \beta \\ -1 & 2 & -3 \end{vmatrix} = \hat{i}(-9 - 2\beta) - \hat{j}(-3 + \beta) + \hat{k}(2 + 3) = (-9 - 2\beta)\hat{i} + (3 - \beta)\hat{j} + 5\hat{k}$.
Given $|\vec{b} \times \vec{c}| = 5\sqrt{3}$,so $|\vec{b} \times \vec{c}|^2 = 75$.
$(-9 - 2\beta)^2 + (3 - \beta)^2 + 5^2 = 75$
$(81 + 36\beta + 4\beta^2) + (9 - 6\beta + \beta^2) + 25 = 75$
$5\beta^2 + 30\beta + 115 = 75$
$5\beta^2 + 30\beta + 40 = 0 \Rightarrow \beta^2 + 6\beta + 8 = 0$.
Solving for $\beta$,$(\beta + 4)(\beta + 2) = 0$,so $\beta = -4$ or $\beta = -2$.
If $\beta = -4$,then $\alpha = 4$. If $\beta = -2$,then $\alpha = 8$.
Now,$|\vec{a}|^2 = 1^2 + 5^2 + \alpha^2 = 26 + \alpha^2$.
For $\alpha = 4$,$|\vec{a}|^2 = 26 + 16 = 42$.
For $\alpha = 8$,$|\vec{a}|^2 = 26 + 64 = 90$.
The greatest value is $90$.

(C) Given that $\vec{a}, \vec{b}, \vec{c}$ are mutually perpendicular vectors with the same magnitude,let $|\vec{a}| = |\vec{b}| = |\vec{c}| = k$.
Using the vector triple product identity $\vec{u} \times (\vec{v} \times \vec{w}) = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{u} \cdot \vec{v})\vec{w}$,we expand each term:
$\vec{a} \times \{(\vec{r}-\vec{b}) \times \vec{a}\} = (\vec{a} \cdot \vec{a})(\vec{r}-\vec{b}) - (\vec{a} \cdot (\vec{r}-\vec{b}))\vec{a} = k^2(\vec{r}-\vec{b}) - (\vec{a} \cdot \vec{r})\vec{a}$.
Similarly,$\vec{b} \times \{(\vec{r}-\vec{c}) \times \vec{b}\} = k^2(\vec{r}-\vec{c}) - (\vec{b} \cdot \vec{r})\vec{b}$ and $\vec{c} \times \{(\vec{r}-\vec{a}) \times \vec{c}\} = k^2(\vec{r}-\vec{a}) - (\vec{c} \cdot \vec{r})\vec{c}$.
Summing these,we get $k^2(3\vec{r} - (\vec{a}+\vec{b}+\vec{c})) - ((\vec{a} \cdot \vec{r})\vec{a} + (\vec{b} \cdot \vec{r})\vec{b} + (\vec{c} \cdot \vec{r})\vec{c}) = \vec{0}$.
Since $\vec{r} = x\vec{a} + y\vec{b} + z\vec{c}$ where $x = \frac{\vec{r} \cdot \vec{a}}{k^2}$,$y = \frac{\vec{r} \cdot \vec{b}}{k^2}$,$z = \frac{\vec{r} \cdot \vec{c}}{k^2}$,the expression becomes $k^2(3\vec{r} - (\vec{a}+\vec{b}+\vec{c})) - k^2(x\vec{a} + y\vec{b} + z\vec{c}) = \vec{0}$.
This simplifies to $3\vec{r} - (\vec{a}+\vec{b}+\vec{c}) - \vec{r} = \vec{0}$,which gives $2\vec{r} = \vec{a}+\vec{b}+\vec{c}$.
Therefore,$\vec{r} = \frac{1}{2}(\vec{a}+\vec{b}+\vec{c})$.

(D) Given $\vec{a} = 2\hat{i} + \hat{j} - 2\hat{k}$,so $|\vec{a}|^2 = 2^2 + 1^2 + (-2)^2 = 4 + 1 + 4 = 9$,which implies $|\vec{a}| = 3$.
Given $|\vec{c} - \vec{a}| = 2\sqrt{2}$. Squaring both sides,we get $|\vec{c}|^2 + |\vec{a}|^2 - 2(\vec{c} \cdot \vec{a}) = (2\sqrt{2})^2 = 8$.
Since $\vec{a} \cdot \vec{c} = |\vec{c}|$,let $|\vec{c}| = c$. Then $c^2 + 9 - 2c = 8$.
$c^2 - 2c + 1 = 0 \Rightarrow (c - 1)^2 = 0 \Rightarrow c = 1$. Thus,$|\vec{c}| = 1$.
Now,calculate $\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 1 & 1 & 0 \end{vmatrix} = \hat{i}(0 - (-2)) - \hat{j}(0 - (-2)) + \hat{k}(2 - 1) = 2\hat{i} - 2\hat{j} + \hat{k}$.
The magnitude $|\vec{a} \times \vec{b}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.
The magnitude of the cross product is $|(\vec{a} \times \vec{b}) \times \vec{c}| = |\vec{a} \times \vec{b}| |\vec{c}| \sin(\theta)$,where $\theta = \frac{\pi}{6}$.
$|(\vec{a} \times \vec{b}) \times \vec{c}| = (3)(1) \sin(\frac{\pi}{6}) = 3 \times \frac{1}{2} = \frac{3}{2}$.

(A) Given vectors are $\vec{p}=2 \hat{i}+3 \hat{j}+\hat{k}$ and $\vec{q}=\hat{i}+2 \hat{j}+\hat{k}$.
First,calculate the sum and difference of the vectors:
$\vec{p}+\vec{q} = (2+1)\hat{i} + (3+2)\hat{j} + (1+1)\hat{k} = 3\hat{i} + 5\hat{j} + 2\hat{k}$
$\vec{p}-\vec{q} = (2-1)\hat{i} + (3-2)\hat{j} + (1-1)\hat{k} = \hat{i} + \hat{j} + 0\hat{k}$
Since $\vec{r}$ is perpendicular to both $(\vec{p}+\vec{q})$ and $(\vec{p}-\vec{q})$,$\vec{r}$ must be parallel to their cross product:
$(\vec{p}+\vec{q}) \times (\vec{p}-\vec{q}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 5 & 2 \\ 1 & 1 & 0 \end{vmatrix} = \hat{i}(0-2) - \hat{j}(0-2) + \hat{k}(3-5) = -2\hat{i} + 2\hat{j} - 2\hat{k}$.
Let $\vec{v} = -2\hat{i} + 2\hat{j} - 2\hat{k}$. The magnitude is $|\vec{v}| = \sqrt{(-2)^2 + 2^2 + (-2)^2} = \sqrt{4+4+4} = \sqrt{12} = 2\sqrt{3}$.
The vector $\vec{r}$ is given by $\vec{r} = \pm |\vec{r}| \frac{\vec{v}}{|\vec{v}|} = \pm \sqrt{3} \frac{-2\hat{i} + 2\hat{j} - 2\hat{k}}{2\sqrt{3}} = \pm (-\hat{i} + \hat{j} - \hat{k})$.
Thus,$\vec{r} = \mp \hat{i} \pm \hat{j} \mp \hat{k}$.
Comparing with $\vec{r} = \alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}$,we get $|\alpha|=1, |\beta|=1, |\gamma|=1$.
Therefore,$|\alpha|+|\beta|+|\gamma| = 1+1+1 = 3$.

(C) Given $\vec{a} \times \vec{b} = 13 \hat{i} - \hat{j} - 4 \hat{k}$. Since $(\vec{a} \times \vec{b}) \cdot \vec{b} = 0$,we have $(13 \hat{i} - \hat{j} - 4 \hat{k}) \cdot (\hat{i} + \hat{j} + \lambda \hat{k}) = 0$.
$13 - 1 - 4\lambda = 0 \Rightarrow 4\lambda = 12 \Rightarrow \lambda = 3$.
Thus,$\vec{b} = \hat{i} + \hat{j} + 3 \hat{k}$.
Using the vector triple product identity $(\vec{a} \times \vec{b}) \times \vec{b} = (\vec{a} \cdot \vec{b}) \vec{b} - (\vec{b} \cdot \vec{b}) \vec{a}$,we have:
$(\vec{a} \times \vec{b}) \times \vec{b} = (13 \hat{i} - \hat{j} - 4 \hat{k}) \times (\hat{i} + \hat{j} + 3 \hat{k}) = \hat{i}(-3+4) - \hat{j}(39+4) + \hat{k}(13+1) = \hat{i} - 43 \hat{j} + 14 \hat{k}$.
Given $\vec{a} \cdot \vec{b} = -21$ and $|\vec{b}|^2 = 1^2 + 1^2 + 3^2 = 11$,we have $-21 \vec{b} - 11 \vec{a} = \hat{i} - 43 \hat{j} + 14 \hat{k}$.
$-21(\hat{i} + \hat{j} + 3 \hat{k}) - 11 \vec{a} = \hat{i} - 43 \hat{j} + 14 \hat{k} \Rightarrow 11 \vec{a} = -22 \hat{i} + 22 \hat{j} - 77 \hat{k} \Rightarrow \vec{a} = -2 \hat{i} + 2 \hat{j} - 7 \hat{k}$.
Now,calculate $(\vec{b}-\vec{a}) \cdot(\hat{k}-\hat{j})+(\vec{b}+\vec{a}) \cdot(\hat{i}-\hat{k})$:
$\vec{b}-\vec{a} = (1 - (-2))\hat{i} + (1 - 2)\hat{j} + (3 - (-7))\hat{k} = 3 \hat{i} - \hat{j} + 10 \hat{k}$.
$\vec{b}+\vec{a} = (1 - 2)\hat{i} + (1 + 2)\hat{j} + (3 - 7)\hat{k} = -\hat{i} + 3 \hat{j} - 4 \hat{k}$.
$(\vec{b}-\vec{a}) \cdot(\hat{k}-\hat{j}) = (3 \hat{i} - \hat{j} + 10 \hat{k}) \cdot (0 \hat{i} - 1 \hat{j} + 1 \hat{k}) = 0 + 1 + 10 = 11$.
$(\vec{b}+\vec{a}) \cdot(\hat{i}-\hat{k}) = (-\hat{i} + 3 \hat{j} - 4 \hat{k}) \cdot (1 \hat{i} + 0 \hat{j} - 1 \hat{k}) = -1 + 0 + 4 = 3$.
Sum $= 11 + 3 = 14$.

(A) Given $|\vec{a}|=4$ and $|\vec{b}|=3$.
We need to evaluate $|(\vec{a}-\vec{b}) \times (\vec{a}+\vec{b})|^{2} + 4(\vec{a} \cdot \vec{b})^{2}$.
Expanding the cross product:
$(\vec{a}-\vec{b}) \times (\vec{a}+\vec{b}) = \vec{a} \times \vec{a} + \vec{a} \times \vec{b} - \vec{b} \times \vec{a} - \vec{b} \times \vec{b}$.
Since $\vec{a} \times \vec{a} = 0$ and $\vec{b} \times \vec{b} = 0$,and $\vec{b} \times \vec{a} = -(\vec{a} \times \vec{b})$,we get:
$(\vec{a}-\vec{b}) \times (\vec{a}+\vec{b}) = 0 + \vec{a} \times \vec{b} + \vec{a} \times \vec{b} - 0 = 2(\vec{a} \times \vec{b})$.
Now,$|2(\vec{a} \times \vec{b})|^{2} = 4|\vec{a} \times \vec{b}|^{2} = 4|\vec{a}|^{2}|\vec{b}|^{2} \sin^{2} \theta$.
Also,$4(\vec{a} \cdot \vec{b})^{2} = 4(|\vec{a}||\vec{b}| \cos \theta)^{2} = 4|\vec{a}|^{2}|\vec{b}|^{2} \cos^{2} \theta$.
Adding these:
$4|\vec{a}|^{2}|\vec{b}|^{2} \sin^{2} \theta + 4|\vec{a}|^{2}|\vec{b}|^{2} \cos^{2} \theta = 4|\vec{a}|^{2}|\vec{b}|^{2} (\sin^{2} \theta + \cos^{2} \theta) = 4|\vec{a}|^{2}|\vec{b}|^{2}$.
Substituting the values $|\vec{a}|=4$ and $|\vec{b}|=3$:
$4 \times (4)^{2} \times (3)^{2} = 4 \times 16 \times 9 = 576$.

(D) The area of a parallelogram with diagonals $\vec{a}$ and $\vec{b}$ is given by $\text{Area} = \frac{1}{2} |\vec{a} \times \vec{b}| = 2 \sqrt{2}$.
Thus,$|\vec{a} \times \vec{b}| = 4 \sqrt{2}$.
Given $|\vec{a}|=1$ and $|\vec{a} \cdot \vec{b}| = |\vec{a} \times \vec{b}|$,we have $|\vec{a}| |\vec{b}| \cos \theta = |\vec{a}| |\vec{b}| \sin \theta$,where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$.
Since $\theta$ is acute,$\cos \theta = \sin \theta \Rightarrow \theta = \frac{\pi}{4}$.
Now,$|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \frac{\pi}{4} = 1 \cdot |\vec{b}| \cdot \frac{1}{\sqrt{2}} = 4 \sqrt{2} \Rightarrow |\vec{b}| = 8$.
Given $\vec{c} = 2 \sqrt{2}(\vec{a} \times \vec{b}) - 2 \vec{b}$.
Since $(\vec{a} \times \vec{b})$ is perpendicular to both $\vec{a}$ and $\vec{b}$,the vectors $(2 \sqrt{2}(\vec{a} \times \vec{b}))$ and $(-2 \vec{b})$ are orthogonal.
Thus,$|\vec{c}|^2 = |2 \sqrt{2}(\vec{a} \times \vec{b})|^2 + |-2 \vec{b}|^2 = 8(4 \sqrt{2})^2 + 4(8)^2 = 8(32) + 4(64) = 256 + 256 = 512$.
$|\vec{c}| = \sqrt{512} = 16 \sqrt{2}$.
Now,$\vec{b} \cdot \vec{c} = \vec{b} \cdot (2 \sqrt{2}(\vec{a} \times \vec{b}) - 2 \vec{b}) = 2 \sqrt{2} (\vec{b} \cdot (\vec{a} \times \vec{b})) - 2 |\vec{b}|^2 = 0 - 2(8)^2 = -128$.
Let $\alpha$ be the angle between $\vec{b}$ and $\vec{c}$. Then $\cos \alpha = \frac{\vec{b} \cdot \vec{c}}{|\vec{b}| |\vec{c}|} = \frac{-128}{8 \cdot 16 \sqrt{2}} = \frac{-128}{128 \sqrt{2}} = -\frac{1}{\sqrt{2}}$.
Therefore,$\alpha = \frac{3 \pi}{4}$.

(D) Given vectors are $\vec{a} = \hat{i} + \hat{j} - \hat{k}$ and $\vec{c} = 2\hat{i} - 3\hat{j} + 2\hat{k}$.
We are given the condition $\vec{b} \times \vec{c} = \vec{a}$.
By the definition of the cross product,the vector $\vec{a}$ must be perpendicular to both $\vec{b}$ and $\vec{c}$.
This implies that $\vec{a} \cdot \vec{c} = 0$.
Let us calculate the dot product $\vec{a} \cdot \vec{c}$:
$\vec{a} \cdot \vec{c} = (1)(2) + (1)(-3) + (-1)(2) = 2 - 3 - 2 = -3$.
Since $\vec{a} \cdot \vec{c} = -3 \neq 0$,the vector $\vec{a}$ is not perpendicular to $\vec{c}$.
Therefore,there exists no vector $\vec{b}$ such that $\vec{b} \times \vec{c} = \vec{a}$.
Thus,the number of such vectors $\vec{b}$ is $0$.

(D) Given $\overrightarrow{a} = \alpha \hat{i} + 2 \hat{j} - \hat{k}$ and $\overrightarrow{b} = -2 \hat{i} + \alpha \hat{j} + \hat{k}$.
The area of the parallelogram is given by $|\vec{a} \times \vec{b}|$.
Calculating the cross product: $\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \alpha & 2 & -1 \\ -2 & \alpha & 1 \end{vmatrix} = \hat{i}(2 + \alpha) - \hat{j}(\alpha - 2) + \hat{k}(\alpha^{2} + 4)$.
$|\vec{a} \times \vec{b}| = \sqrt{(\alpha + 2)^{2} + (\alpha - 2)^{2} + (\alpha^{2} + 4)^{2}} = \sqrt{\alpha^{2} + 4\alpha + 4 + \alpha^{2} - 4\alpha + 4 + (\alpha^{2} + 4)^{2}} = \sqrt{2(\alpha^{2} + 4) + (\alpha^{2} + 4)^{2}}$.
Given $|\vec{a} \times \vec{b}| = \sqrt{15(\alpha^{2} + 4)}$,we have $2(\alpha^{2} + 4) + (\alpha^{2} + 4)^{2} = 15(\alpha^{2} + 4)$.
Dividing by $(\alpha^{2} + 4)$,we get $2 + (\alpha^{2} + 4) = 15$,so $\alpha^{2} + 4 = 13$,which means $\alpha^{2} = 9$.
Now,calculate $|\vec{a}|^{2} = \alpha^{2} + 2^{2} + (-1)^{2} = \alpha^{2} + 5 = 9 + 5 = 14$.
$|\vec{b}|^{2} = (-2)^{2} + \alpha^{2} + 1^{2} = 4 + \alpha^{2} + 1 = \alpha^{2} + 5 = 14$.
$\vec{a} \cdot \vec{b} = \alpha(-2) + 2(\alpha) + (-1)(1) = -2\alpha + 2\alpha - 1 = -1$.
Finally,$2|\vec{a}|^{2} + (\vec{a} \cdot \vec{b})|\vec{b}|^{2} = 2(14) + (-1)(14) = 28 - 14 = 14$.

(A) The projection of $\overrightarrow{a}$ on $\overrightarrow{c}$ is given by $\frac{\overrightarrow{a} \cdot \overrightarrow{c}}{|\overrightarrow{c}|} = \frac{10}{3}$.
Calculating $\overrightarrow{a} \cdot \overrightarrow{c} = (\alpha)(1) + (3)(2) + (-1)(-2) = \alpha + 6 + 2 = \alpha + 8$.
Calculating $|\overrightarrow{c}| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.
Thus,$\frac{\alpha + 8}{3} = \frac{10}{3} \Rightarrow \alpha + 8 = 10 \Rightarrow \alpha = 2$.
Next,we calculate $\overrightarrow{b} \times \overrightarrow{c}$:
$\overrightarrow{b} \times \overrightarrow{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -\beta & 4 \\ 1 & 2 & -2 \end{vmatrix} = \hat{i}(2\beta - 8) - \hat{j}(-6 - 4) + \hat{k}(6 + \beta) = (2\beta - 8)\hat{i} + 10\hat{j} + (6 + \beta)\hat{k}$.
Given $\overrightarrow{b} \times \overrightarrow{c} = -6 \hat{i} + 10 \hat{j} + 7 \hat{k}$,comparing the components:
$2\beta - 8 = -6 \Rightarrow 2\beta = 2 \Rightarrow \beta = 1$.
Therefore,$\alpha + \beta = 2 + 1 = 3$.

(A) The normal to the first plane is $\vec{n}_{1} = \hat{i} \times (\hat{i} + \hat{j}) = \hat{k}$.
The normal to the second plane is $\vec{n}_{2} = (\hat{i} - \hat{j}) \times (\hat{i} + \hat{k}) = \hat{i} + \hat{j} + \hat{k}$.
The line of intersection is parallel to $\vec{v} = \vec{n}_{1} \times \vec{n}_{2} = \hat{k} \times (\hat{i} + \hat{j} + \hat{k}) = -\hat{i} + \hat{j}$.
Thus,$\vec{a} = -\hat{i} + \hat{j}$.
Given $\vec{b} = \hat{i} - 2\hat{j} + 2\hat{k}$.
The cosine of the angle $\theta$ between $\vec{a}$ and $\vec{b}$ is given by $\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$.
$\vec{a} \cdot \vec{b} = (-1)(1) + (1)(-2) + (0)(2) = -1 - 2 = -3$.
$|\vec{a}| = \sqrt{(-1)^2 + 1^2} = \sqrt{2}$.
$|\vec{b}| = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{1 + 4 + 4} = 3$.
$\cos \theta = \frac{-3}{\sqrt{2} \times 3} = -\frac{1}{\sqrt{2}}$.
Since $\cos \theta = -\frac{1}{\sqrt{2}}$,the obtuse angle is $\theta = \frac{3\pi}{4}$.

(D) First,calculate the cross product $\vec{a} \times \vec{b}$:
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \alpha & 1 & -1 \\ 2 & 1 & -\alpha \end{vmatrix} = \hat{i}(-\alpha + 1) - \hat{j}(-\alpha^2 + 2) + \hat{k}(\alpha - 2) = (1 - \alpha) \hat{i} + (\alpha^2 - 2) \hat{j} + (\alpha - 2) \hat{k}$.
Let $\vec{v} = \vec{a} \times \vec{b}$ and $\vec{c} = -\hat{i} + 2 \hat{j} - 2 \hat{k}$. The magnitude of $\vec{c}$ is $|\vec{c}| = \sqrt{(-1)^2 + 2^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.
The projection of $\vec{v}$ on $\vec{c}$ is given by $\frac{\vec{v} \cdot \vec{c}}{|\vec{c}|} = 30$.
$\vec{v} \cdot \vec{c} = (1 - \alpha)(-1) + (\alpha^2 - 2)(2) + (\alpha - 2)(-2) = -1 + \alpha + 2\alpha^2 - 4 - 2\alpha + 4 = 2\alpha^2 - \alpha - 1$.
So,$\frac{2\alpha^2 - \alpha - 1}{3} = 30 \implies 2\alpha^2 - \alpha - 1 = 90 \implies 2\alpha^2 - \alpha - 91 = 0$.
Solving the quadratic equation $2\alpha^2 - 14\alpha + 13\alpha - 91 = 0 \implies 2\alpha(\alpha - 7) + 13(\alpha - 7) = 0$.
Thus,$(\alpha - 7)(2\alpha + 13) = 0$,which gives $\alpha = 7$ or $\alpha = -\frac{13}{2}$.
Since $\alpha > 0$,we have $\alpha = 7$.

(D) Given $\vec{a} \times \vec{b} = 4\vec{c}$,$\vec{b} \times \vec{c} = 9\vec{a}$,and $\vec{c} \times \vec{a} = \alpha\vec{b}$.
Taking the magnitude of the first equation: $|\vec{a} \times \vec{b}| = |4\vec{c}| \implies |\vec{a}||\vec{b}| \sin \theta_1 = 4|\vec{c}|$. Since $\vec{a}, \vec{b}, \vec{c}$ are mutually orthogonal (as $\vec{a} \times \vec{b}$ is parallel to $\vec{c}$),we have $|\vec{a}||\vec{b}| = 4|\vec{c}|$.
Similarly,$|\vec{b}||\vec{c}| = 9|\vec{a}|$ and $|\vec{c}||\vec{a}| = \alpha|\vec{b}|$.
Multiplying these three equations: $(|\vec{a}||\vec{b}||\vec{c}|)^2 = 4 \times 9 \times \alpha \times |\vec{a}||\vec{b}||\vec{c}|$.
Thus,$|\vec{a}||\vec{b}||\vec{c}| = 36\alpha$.
From $|\vec{a}||\vec{b}| = 4|\vec{c}|$,we get $|\vec{a}||\vec{b}||\vec{c}| = 4|\vec{c}|^2 = 36\alpha \implies |\vec{c}|^2 = 9\alpha \implies |\vec{c}| = 3\sqrt{\alpha}$.
Similarly,$|\vec{a}| = 2\sqrt{\alpha}$ and $|\vec{b}| = 6$ (since $|\vec{b}|^2 = 36 \implies |\vec{b}| = 6$ is not correct,let's re-evaluate).
Using $|\vec{a}| = 2\sqrt{\alpha}, |\vec{b}| = 3\sqrt{\alpha}, |\vec{c}| = 6$ is wrong. Let's use $|\vec{a}||\vec{b}| = 4|\vec{c}|, |\vec{b}||\vec{c}| = 9|\vec{a}|, |\vec{c}||\vec{a}| = \alpha|\vec{b}|$.
Dividing equations: $\frac{|\vec{a}|}{|\vec{c}|} = \frac{4|\vec{c}|}{9|\vec{a}|} \implies 9|\vec{a}|^2 = 4|\vec{c}|^2 \implies 3|\vec{a}| = 2|\vec{c}|$.
Also $\frac{|\vec{b}|}{|\vec{a}|} = \frac{9|\vec{a}|}{\alpha|\vec{b}|} \implies \alpha|\vec{b}|^2 = 9|\vec{a}|^2$.
Given $|\vec{a}| + |\vec{b}| + |\vec{c}| = 36$. Let $|\vec{a}| = 2k, |\vec{c}| = 3k$. Then $4|\vec{c}| = 12k = |\vec{a}||\vec{b}| = 2k|\vec{b}| \implies |\vec{b}| = 6$.
Then $9|\vec{a}|^2 = \alpha|\vec{b}|^2 \implies 9(4k^2) = \alpha(36) \implies 36k^2 = 36\alpha \implies k^2 = \alpha \implies k = \sqrt{\alpha}$.
Substituting into sum: $2\sqrt{\alpha} + 6 + 3\sqrt{\alpha} = 36 \implies 5\sqrt{\alpha} = 30 \implies \sqrt{\alpha} = 6 \implies \alpha = 36$.

(B) Given $|\vec{a}| = 9$. Since $(x \vec{a} + y \vec{b}) \perp (6y \vec{a} - 18x \vec{b})$,their dot product is zero:
$(x \vec{a} + y \vec{b}) \cdot (6y \vec{a} - 18x \vec{b}) = 0$
$6xy |\vec{a}|^2 - 18x^2 (\vec{a} \cdot \vec{b}) + 6y^2 (\vec{a} \cdot \vec{b}) - 18xy |\vec{b}|^2 = 0$
$6xy (|\vec{a}|^2 - 3|\vec{b}|^2) + (\vec{a} \cdot \vec{b})(6y^2 - 18x^2) = 0$
For this to hold for all $(x, y)$,the coefficients of $xy$,$x^2$,and $y^2$ must be zero:
$|\vec{a}|^2 - 3|\vec{b}|^2 = 0 \implies |\vec{b}|^2 = \frac{|\vec{a}|^2}{3} = \frac{81}{3} = 27$
$\vec{a} \cdot \vec{b} = 0$
Now,$|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2 = 81 \times 27 - 0 = 2187$
$|\vec{a} \times \vec{b}| = \sqrt{2187} = \sqrt{81 \times 27} = 9 \times 3 \sqrt{3} = 27 \sqrt{3}$.

(D) Given vectors are $\vec{u}=2 \hat{i}-\hat{j}+\hat{k}$ and $\vec{v}=-3 \hat{j}+2 \hat{k}$.
First,calculate the cross product $\vec{u} \times \vec{v}$:
$\vec{u} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 0 & -3 & 2 \end{vmatrix}$
$= \hat{i}(-2 - (-3)) - \hat{j}(4 - 0) + \hat{k}(-6 - 0)$
$= \hat{i}(1) - 4 \hat{j} - 6 \hat{k} = \hat{i} - 4 \hat{j} - 6 \hat{k}$.
Since $\vec{w}$ is a unit vector in the $XY$-plane,it can be represented as $\vec{w} = \cos \theta \hat{i} + \sin \theta \hat{j}$.
Now,calculate the dot product $|(\vec{u} \times \vec{v}) \cdot \vec{w}|$:
$|(\hat{i} - 4 \hat{j} - 6 \hat{k}) \cdot (\cos \theta \hat{i} + \sin \theta \hat{j})| = |\cos \theta - 4 \sin \theta|$.
The maximum value of an expression of the form $a \cos \theta + b \sin \theta$ is $\sqrt{a^2 + b^2}$.
Here,$a = 1$ and $b = -4$.
Maximum value $= \sqrt{1^2 + (-4)^2} = \sqrt{1 + 16} = \sqrt{17}$.

(D) Given $\vec{a}=\hat{i}+2 \hat{j}+\lambda \hat{k}$ and $\vec{b}=3 \hat{i}-5 \hat{j}-\lambda \hat{k}$.
From $\vec{a} \times \vec{c}=\vec{b} \times \vec{c}$,we have $(\vec{a}-\vec{b}) \times \vec{c} = \vec{0}$.
This implies that $(\vec{a}-\vec{b})$ is parallel to $\vec{c}$,so $\vec{a}-\vec{b} = \mu \vec{c}$ for some scalar $\mu$.
Calculating $\vec{a}-\vec{b} = (1-3)\hat{i} + (2-(-5))\hat{j} + (\lambda - (-\lambda))\hat{k} = -2\hat{i} + 7\hat{j} + 2\lambda\hat{k}$.
Thus,$\mu \vec{c} = -2\hat{i} + 7\hat{j} + 2\lambda\hat{k}$.
Given $\vec{a} \cdot \vec{c} = 7$,we have $\vec{a} \cdot (\frac{1}{\mu} (-2\hat{i} + 7\hat{j} + 2\lambda\hat{k})) = 7$,which gives $-2 + 14 + 2\lambda^2 = 7\mu$,so $12 + 2\lambda^2 = 7\mu$.
Given $2\vec{b} \cdot \vec{c} = -43$,we have $\vec{b} \cdot (\frac{1}{\mu} (-2\hat{i} + 7\hat{j} + 2\lambda\hat{k})) = -\frac{43}{2}$,which gives $-6 - 35 - 2\lambda^2 = -\frac{43}{2}\mu$,so $41 + 2\lambda^2 = \frac{43}{2}\mu$.
Solving the system $2\lambda^2 = 7\mu - 12$ and $2\lambda^2 = \frac{43}{2}\mu - 41$,we get $7\mu - 12 = 21.5\mu - 41$,so $14.5\mu = 29$,which gives $\mu = 2$.
Then $2\lambda^2 = 7(2) - 12 = 2$,so $\lambda^2 = 1$.
Finally,$\vec{a} \cdot \vec{b} = (1)(3) + (2)(-5) + (\lambda)(-\lambda) = 3 - 10 - \lambda^2 = -7 - 1 = -8$.
Thus,$|\vec{a} \cdot \vec{b}| = |-8| = 8$.

(A) Given $\vec{a} = -\hat{i} + 2\hat{j} + \hat{k}$. The vector $\vec{b}$ is obtained by rotating $\vec{a}$ by $90^{\circ}$ such that it passes through the $y$-axis. This implies $\vec{b}$ lies in the plane of $\vec{a}$ and $\hat{j}$.
Thus,$\vec{b} = \lambda(\vec{a} \times (\vec{a} \times \hat{j}))$.
Calculating the triple product: $\vec{a} \times \hat{j} = (-\hat{i} + 2\hat{j} + \hat{k}) \times \hat{j} = -\hat{k} + \hat{i} = \hat{i} - \hat{k}$.
Then $\vec{a} \times (\vec{a} \times \hat{j}) = (-\hat{i} + 2\hat{j} + \hat{k}) \times (\hat{i} - \hat{k}) = \hat{j} + 2\hat{k} + 2\hat{i} + \hat{j} = 2\hat{i} + 2\hat{j} + 2\hat{k}$.
Since $|\vec{b}| = |\vec{a}| = \sqrt{(-1)^2 + 2^2 + 1^2} = \sqrt{6}$,we have $\sqrt{6} = |\lambda| \sqrt{2^2 + 2^2 + 2^2} = |\lambda| \sqrt{12} = 2\sqrt{3}|\lambda|$.
So,$|\lambda| = \frac{\sqrt{6}}{2\sqrt{3}} = \frac{1}{\sqrt{2}}$.
Since $\vec{b}$ passes through the $y$-axis,$\vec{b} \cdot \hat{j} > 0$. Testing $\lambda = -\frac{1}{\sqrt{2}}$,$\vec{b} = -\frac{1}{\sqrt{2}}(2\hat{i} + 2\hat{j} + 2\hat{k}) = -\sqrt{2}\hat{i} - \sqrt{2}\hat{j} - \sqrt{2}\hat{k}$. This gives $\vec{b} \cdot \hat{j} = -\sqrt{2} < 0$.
Testing $\lambda = \frac{1}{\sqrt{2}}$,$\vec{b} = \sqrt{2}\hat{i} + \sqrt{2}\hat{j} + \sqrt{2}\hat{k}$. This gives $\vec{b} \cdot \hat{j} = \sqrt{2} > 0$. Thus $\vec{b} = \sqrt{2}\hat{i} + \sqrt{2}\hat{j} + \sqrt{2}\hat{k}$.
Now,$3\vec{a} + \sqrt{2}\vec{b} = 3(-\hat{i} + 2\hat{j} + \hat{k}) + \sqrt{2}(\sqrt{2}\hat{i} + \sqrt{2}\hat{j} + \sqrt{2}\hat{k}) = -3\hat{i} + 6\hat{j} + 3\hat{k} + 2\hat{i} + 2\hat{j} + 2\hat{k} = -\hat{i} + 8\hat{j} + 5\hat{k}$.
The projection on $\vec{c} = 5\hat{i} + 4\hat{j} + 3\hat{k}$ is $\frac{(- \hat{i} + 8\hat{j} + 5\hat{k}) \cdot (5\hat{i} + 4\hat{j} + 3\hat{k})}{\sqrt{5^2 + 4^2 + 3^2}} = \frac{-5 + 32 + 15}{\sqrt{50}} = \frac{42}{5\sqrt{2}} = \frac{21\sqrt{2}}{5} = 4.2\sqrt{2}$.

(A) Given the expression $((\vec{a} + \vec{b}) \times (\vec{a} \times \vec{b})) \times (\vec{a} - \vec{b}) = 8 \hat{i} - 40 \hat{j} - 24 \hat{k}$.
Using the vector triple product identity $(\vec{u} \times \vec{v}) \times \vec{w} = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{v} \cdot \vec{w})\vec{u}$,let $\vec{u} = \vec{a} + \vec{b}$,$\vec{v} = \vec{a} \times \vec{b}$,and $\vec{w} = \vec{a} - \vec{b}$.
Since $(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = |\vec{a}|^2 - |\vec{b}|^2 = (\lambda^2 + 4 + 9) - (1 + \lambda^2 + 4) = 8$,the expression simplifies to $8(\vec{a} \times \vec{b}) = 8 \hat{i} - 40 \hat{j} - 24 \hat{k}$.
Thus,$\vec{a} \times \vec{b} = \hat{i} - 5 \hat{j} - 3 \hat{k}$.
Calculating $\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \lambda & 2 & -3 \\ 1 & -\lambda & 2 \end{vmatrix} = (4 - 3\lambda)\hat{i} - (2\lambda + 3)\hat{j} + (-\lambda^2 - 2)\hat{k}$.
Comparing components: $4 - 3\lambda = 1 \Rightarrow \lambda = 1$. Check: $-(2(1) + 3) = -5$ and $-(1^2 + 2) = -3$. This matches.
For $\lambda = 1$,$\vec{a} = \hat{i} + 2\hat{j} - 3\hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + 2\hat{k}$.
Then $\vec{a} + \vec{b} = 2\hat{i} + \hat{j} - \hat{k}$ and $\vec{a} - \vec{b} = 3\hat{j} - 5\hat{k}$.
$(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ 0 & 3 & -5 \end{vmatrix} = (-5 + 3)\hat{i} - (-10 - 0)\hat{j} + (6 - 0)\hat{k} = -2\hat{i} + 10\hat{j} + 6\hat{k}$.
Since $\lambda = 1$,we need $|1(-2\hat{i} + 10\hat{j} + 6\hat{k})|^2 = (-2)^2 + 10^2 + 6^2 = 4 + 100 + 36 = 140$.

(D) Given $2(\vec{a} \times \vec{b}) = 3(\vec{c} \times \vec{a})$.
This can be written as $\vec{a} \times (2\vec{b} + 3\vec{c}) = 0$.
This implies $\vec{a} = \lambda(2\vec{b} + 3\vec{c})$ for some scalar $\lambda$.
Given $|\vec{a}| = \sqrt{31}$,$|\vec{b}| = 1/2$,and $|\vec{c}| = 2$.
The angle between $\vec{b}$ and $\vec{c}$ is $\theta = \frac{2\pi}{3}$,so $\vec{b} \cdot \vec{c} = |\vec{b}||\vec{c}| \cos(2\pi/3) = (1/2)(2)(-1/2) = -1/2$.
Now,$|\vec{a}|^2 = \lambda^2 |2\vec{b} + 3\vec{c}|^2 = \lambda^2 (4|\vec{b}|^2 + 9|\vec{c}|^2 + 12\vec{b} \cdot \vec{c})$.
$31 = \lambda^2 (4(1/4) + 9(4) + 12(-1/2)) = \lambda^2 (1 + 36 - 6) = 31\lambda^2$.
Thus,$\lambda^2 = 1$,so $\lambda = \pm 1$.
Then $\vec{a} = \pm(2\vec{b} + 3\vec{c})$.
We need to evaluate $\left(\frac{\vec{a} \times \vec{c}}{\vec{a} \cdot \vec{b}}\right)^2 = \frac{|\vec{a} \times \vec{c}|^2}{(\vec{a} \cdot \vec{b})^2}$.
$|\vec{a} \times \vec{c}|^2 = |\pm(2\vec{b} + 3\vec{c}) \times \vec{c}|^2 = |2(\vec{b} \times \vec{c})|^2 = 4|\vec{b} \times \vec{c}|^2 = 4(|\vec{b}|^2|\vec{c}|^2 - (\vec{b} \cdot \vec{c})^2) = 4(1/4 \cdot 4 - (-1/2)^2) = 4(1 - 1/4) = 3$.
$\vec{a} \cdot \vec{b} = \pm(2\vec{b} + 3\vec{c}) \cdot \vec{b} = \pm(2|\vec{b}|^2 + 3\vec{b} \cdot \vec{c}) = \pm(2(1/4) + 3(-1/2)) = \pm(1/2 - 3/2) = \pm(-1) = \mp 1$.
So,$(\vec{a} \cdot \vec{b})^2 = 1$.
Therefore,$\left(\frac{\vec{a} \times \vec{c}}{\vec{a} \cdot \vec{b}}\right)^2 = \frac{3}{1} = 3$.

(C) The area of a quadrilateral with vertices $A, B, C, D$ is given by $\text{Area} = \frac{1}{2} |\overrightarrow{AC} \times \overrightarrow{BD}| = 18$,so $|\overrightarrow{AC} \times \overrightarrow{BD}| = 36$.
Given vertices: $A(2,6,2), B(-4,0,\lambda), C(2,3,-1), D(4,5,0)$.
Calculate vectors $\overrightarrow{AC}$ and $\overrightarrow{BD}$:
$\overrightarrow{AC} = (2-2)\hat{i} + (3-6)\hat{j} + (-1-2)\hat{k} = 0\hat{i} - 3\hat{j} - 3\hat{k}$.
$\overrightarrow{BD} = (4-(-4))\hat{i} + (5-0)\hat{j} + (0-\lambda)\hat{k} = 8\hat{i} + 5\hat{j} - \lambda\hat{k}$.
Calculate the cross product $\overrightarrow{AC} \times \overrightarrow{BD}$:
$\overrightarrow{AC} \times \overrightarrow{BD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & -3 & -3 \\ 8 & 5 & -\lambda \end{vmatrix} = \hat{i}(3\lambda + 15) - \hat{j}(0 - (-24)) + \hat{k}(0 - (-24)) = (3\lambda + 15)\hat{i} - 24\hat{j} + 24\hat{k}$.
Calculate the magnitude:
$|\overrightarrow{AC} \times \overrightarrow{BD}| = \sqrt{(3\lambda + 15)^2 + (-24)^2 + (24)^2} = 36$.
Squaring both sides: $(3\lambda + 15)^2 + 576 + 576 = 1296$.
$(3\lambda + 15)^2 = 1296 - 1152 = 144$.
$3\lambda + 15 = \pm 12$.
Case $1$: $3\lambda + 15 = 12 \Rightarrow 3\lambda = -3 \Rightarrow \lambda = -1$.
Case $2$: $3\lambda + 15 = -12 \Rightarrow 3\lambda = -27 \Rightarrow \lambda = -9$.
Since $|\lambda| \leq 5$,we choose $\lambda = -1$.
Finally,$5 - 6\lambda = 5 - 6(-1) = 5 + 6 = 11$.

(B) The area of a quadrilateral $ABCD$ is given by the formula: $\text{Area} = \frac{1}{2} |\overrightarrow{AC} \times \overrightarrow{BD}|$.
First,we find the vectors $\overrightarrow{AC}$ and $\overrightarrow{BD}$:
$\overrightarrow{AC} = C - A = (-2-2, -3-1, 5-1) = (-4, -4, 4) = -4\hat{i} - 4\hat{j} + 4\hat{k}$.
$\overrightarrow{BD} = D - B = (1-1, -6-2, -7-5) = (0, -8, -12) = 0\hat{i} - 8\hat{j} - 12\hat{k}$.
Now,calculate the cross product $\overrightarrow{AC} \times \overrightarrow{BD}$:
$\overrightarrow{AC} \times \overrightarrow{BD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -4 & -4 & 4 \\ 0 & -8 & -12 \end{vmatrix}$
$= \hat{i}((-4)(-12) - (4)(-8)) - \hat{j}((-4)(-12) - (4)(0)) + \hat{k}((-4)(-8) - (-4)(0))$
$= \hat{i}(48 + 32) - \hat{j}(48 - 0) + \hat{k}(32 - 0)$
$= 80\hat{i} - 48\hat{j} + 32\hat{k}$.
Now,find the magnitude of the cross product:
$|\overrightarrow{AC} \times \overrightarrow{BD}| = \sqrt{80^2 + (-48)^2 + 32^2} = \sqrt{6400 + 2304 + 1024} = \sqrt{9728}$.
$\sqrt{9728} = \sqrt{256 \times 38} = 16\sqrt{38}$.
Finally,the area is $\frac{1}{2} |\overrightarrow{AC} \times \overrightarrow{BD}| = \frac{1}{2} \times 16\sqrt{38} = 8\sqrt{38}$.

(A) Given $\overline{OP} = -\hat{i}-2\hat{j}+3\hat{k}$.
$\overline{AB} = \overline{OB} - \overline{OA} = (2\hat{i}+4\hat{j}-2\hat{k}) - (-2\hat{i}+\hat{j}-3\hat{k}) = 4\hat{i}+3\hat{j}+\hat{k}$.
$\overline{AC} = \overline{OC} - \overline{OA} = (-4\hat{i}+2\hat{j}-\hat{k}) - (-2\hat{i}+\hat{j}-3\hat{k}) = -2\hat{i}+\hat{j}+2\hat{k}$.
$A$ vector perpendicular to both $\overline{AB}$ and $\overline{AC}$ is $\vec{n} = \overline{AB} \times \overline{AC}$.
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 3 & 1 \\ -2 & 1 & 2 \end{vmatrix} = \hat{i}(6-1) - \hat{j}(8+2) + \hat{k}(4+6) = 5\hat{i}-10\hat{j}+10\hat{k}$.
The projection of $\overline{OP}$ on $\vec{n}$ is given by $\frac{|\overline{OP} \cdot \vec{n}|}{|\vec{n}|}$.
$\overline{OP} \cdot \vec{n} = (-1)(5) + (-2)(-10) + (3)(10) = -5 + 20 + 30 = 45$.
$|\vec{n}| = \sqrt{5^2 + (-10)^2 + 10^2} = \sqrt{25 + 100 + 100} = \sqrt{225} = 15$.
Projection $= \frac{|45|}{15} = 3$.