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

(D) The normal vectors to the two planes are $\vec{n}_1 = (\hat{i}+\hat{j}) \times (\hat{i}+\hat{k}) = \hat{i}-\hat{j}-\hat{k}$ and $\vec{n}_2 = (\hat{i}-\hat{j}) \times (\hat{j}-\hat{k}) = \hat{i}+\hat{j}+\hat{k}$.
Since $\vec{a}$ is parallel to the line of intersection,$\vec{a} = \lambda(\vec{n}_1 \times \vec{n}_2)$.
$\vec{n}_1 \times \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & -1 \\ 1 & 1 & 1 \end{vmatrix} = 0\hat{i} - 2\hat{j} + 2\hat{k} = -2\hat{j} + 2\hat{k}$.
So,$\vec{a} = \lambda(-2\hat{j} + 2\hat{k})$.
Given $\vec{a} \cdot \vec{b} = 6$,where $\vec{b} = 2\hat{i}-2\hat{j}+\hat{k}$.
$\lambda(-2\hat{j} + 2\hat{k}) \cdot (2\hat{i}-2\hat{j}+\hat{k}) = \lambda(0 + 4 + 2) = 6\lambda = 6 \implies \lambda = 1$.
Thus,$\vec{a} = -2\hat{j} + 2\hat{k}$.
$|\vec{a}| = \sqrt{(-2)^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$ and $|\vec{b}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{9} = 3$.
$\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \frac{6}{2\sqrt{2} \times 3} = \frac{1}{\sqrt{2}} \implies \theta = \frac{\pi}{4}$.
Using the identity $|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2|\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2 = (8)(9) - 6^2 = 72 - 36 = 36$.
Therefore,$|\vec{a} \times \vec{b}| = 6$.
The ordered pair is $(\frac{\pi}{4}, 6)$.

(C) Given $\vec{a} = 3\hat{i} + \hat{j} - \hat{k}$ and $\vec{c} = 2\hat{i} - 3\hat{j} + 3\hat{k}$.
$|\vec{a}|^2 = 3^2 + 1^2 + (-1)^2 = 9 + 1 + 1 = 11$,so $|\vec{a}| = \sqrt{11}$.
$|\vec{c}|^2 = 2^2 + (-3)^2 + 3^2 = 4 + 9 + 9 = 22$,so $|\vec{c}| = \sqrt{22}$.
Since $\vec{a} = \vec{b} \times \vec{c}$,$\vec{a}$ is perpendicular to $\vec{c}$,so $\vec{a} \cdot \vec{c} = 0$.
Also,$|\vec{a}| = |\vec{b} \times \vec{c}| = |\vec{b}||\vec{c}| \sin \theta$,where $\theta$ is the angle between $\vec{b}$ and $\vec{c}$.
$\sqrt{11} = \sqrt{50} \cdot \sqrt{22} \sin \theta \Rightarrow \sqrt{11} = 5\sqrt{2} \cdot \sqrt{22} \sin \theta = 5 \cdot \sqrt{44} \sin \theta = 10\sqrt{11} \sin \theta$.
Thus,$\sin \theta = \frac{1}{10}$.
Then $\cos^2 \theta = 1 - \sin^2 \theta = 1 - \frac{1}{100} = \frac{99}{100}$,so $\cos \theta = \frac{\sqrt{99}}{10}$.
Now,$|\vec{b} + \vec{c}|^2 = |\vec{b}|^2 + |\vec{c}|^2 + 2\vec{b} \cdot \vec{c} = |\vec{b}|^2 + |\vec{c}|^2 + 2|\vec{b}||\vec{c}| \cos \theta$.
$|\vec{b} + \vec{c}|^2 = 50 + 22 + 2(\sqrt{50})(\sqrt{22}) \left(\frac{\sqrt{99}}{10}\right) = 72 + 2(5\sqrt{2})(\sqrt{22}) \left(\frac{3\sqrt{11}}{10}\right) = 72 + 2(5 \cdot 2 \cdot \sqrt{11}) \left(\frac{3\sqrt{11}}{10}\right) = 72 + 20\sqrt{11} \cdot \frac{3\sqrt{11}}{10} = 72 + 2(3 \cdot 11) = 72 + 66 = 138$.
Finally,$|72 - |\vec{b} + \vec{c}|^2| = |72 - 138| = |-66| = 66$.

(D) Given: $|\vec{a}|=2, |\vec{b}|=3$ and the angle $\theta = \frac{\pi}{4}$.
We need to evaluate $|(\vec{a}+2 \vec{b}) \times(2 \vec{a}-3 \vec{b})|^2$.
Expanding the cross product:
$(\vec{a}+2 \vec{b}) \times(2 \vec{a}-3 \vec{b}) = \vec{a} \times (2 \vec{a}) - \vec{a} \times (3 \vec{b}) + (2 \vec{b}) \times (2 \vec{a}) - (2 \vec{b}) \times (3 \vec{b})$.
Since $\vec{a} \times \vec{a} = 0$ and $\vec{b} \times \vec{b} = 0$,this simplifies to:
$0 - 3(\vec{a} \times \vec{b}) + 4(\vec{b} \times \vec{a}) - 0$.
Using the property $\vec{b} \times \vec{a} = -(\vec{a} \times \vec{b})$:
$-3(\vec{a} \times \vec{b}) - 4(\vec{a} \times \vec{b}) = -7(\vec{a} \times \vec{b})$.
Now,calculate the magnitude squared:
$|-7(\vec{a} \times \vec{b})|^2 = 49 |\vec{a} \times \vec{b}|^2$.
Since $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin(\theta)$:
$|\vec{a} \times \vec{b}| = 2 \times 3 \times \sin\left(\frac{\pi}{4}\right) = 6 \times \frac{1}{\sqrt{2}} = 3\sqrt{2}$.
Therefore,$|\vec{a} \times \vec{b}|^2 = (3\sqrt{2})^2 = 9 \times 2 = 18$.
Finally,$49 \times 18 = 882$.

(D) The area of the parallelogram $S$ formed by vectors $\overrightarrow{OA}=\overrightarrow{a}$ and $\overrightarrow{OC}=\overrightarrow{b}$ is given by $|\overrightarrow{a} \times \overrightarrow{b}|$.
The area of the quadrilateral $OABC$ can be calculated as the sum of the areas of $\triangle OAB$ and $\triangle OBC$.
Area of $\triangle OAB = \frac{1}{2} |\overrightarrow{OA} \times \overrightarrow{OB}| = \frac{1}{2} |\overrightarrow{a} \times (12 \overrightarrow{a} + 4 \overrightarrow{b})| = \frac{1}{2} |12(\overrightarrow{a} \times \overrightarrow{a}) + 4(\overrightarrow{a} \times \overrightarrow{b})| = \frac{1}{2} |0 + 4(\overrightarrow{a} \times \overrightarrow{b})| = 2 |\overrightarrow{a} \times \overrightarrow{b}|$.
Area of $\triangle OBC = \frac{1}{2} |\overrightarrow{OC} \times \overrightarrow{OB}| = \frac{1}{2} |\overrightarrow{b} \times (12 \overrightarrow{a} + 4 \overrightarrow{b})| = \frac{1}{2} |12(\overrightarrow{b} \times \overrightarrow{a}) + 4(\overrightarrow{b} \times \overrightarrow{b})| = \frac{1}{2} |12(\overrightarrow{b} \times \overrightarrow{a}) + 0| = 6 |\overrightarrow{b} \times \overrightarrow{a}| = 6 |\overrightarrow{a} \times \overrightarrow{b}|$.
Total area of quadrilateral $OABC = 2 |\overrightarrow{a} \times \overrightarrow{b}| + 6 |\overrightarrow{a} \times \overrightarrow{b}| = 8 |\overrightarrow{a} \times \overrightarrow{b}|$.
The ratio of the area of the quadrilateral $OABC$ to the area of $S$ is $\frac{8 |\overrightarrow{a} \times \overrightarrow{b}|}{|\overrightarrow{a} \times \overrightarrow{b}|} = 8$.

(C) Given $|\vec{a}|=1$,$|\vec{b}|=4$,and $\vec{a} \cdot \vec{b}=2$.
We are given $\vec{c}=2(\vec{a} \times \vec{b})-3 \vec{b}$.
To find the angle $\theta$ between $\vec{b}$ and $\vec{c}$,we use the formula $\cos \theta = \frac{\vec{b} \cdot \vec{c}}{|\vec{b}| |\vec{c}|}$.
First,calculate $\vec{b} \cdot \vec{c}$:
$\vec{b} \cdot \vec{c} = \vec{b} \cdot (2(\vec{a} \times \vec{b}) - 3 \vec{b}) = 2(\vec{b} \cdot (\vec{a} \times \vec{b})) - 3(\vec{b} \cdot \vec{b})$.
Since $\vec{b} \cdot (\vec{a} \times \vec{b}) = 0$ (as the cross product is perpendicular to both vectors),we have $\vec{b} \cdot \vec{c} = 0 - 3|\vec{b}|^2 = -3(4^2) = -3(16) = -48$.
Next,calculate $|\vec{c}|^2$:
$|\vec{c}|^2 = |2(\vec{a} \times \vec{b}) - 3 \vec{b}|^2 = 4|\vec{a} \times \vec{b}|^2 + 9|\vec{b}|^2 - 12(\vec{a} \times \vec{b}) \cdot \vec{b}$.
Since $(\vec{a} \times \vec{b}) \cdot \vec{b} = 0$,we have $|\vec{c}|^2 = 4|\vec{a} \times \vec{b}|^2 + 9|\vec{b}|^2$.
Using $|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2 = (1)^2(4)^2 - (2)^2 = 16 - 4 = 12$.
Thus,$|\vec{c}|^2 = 4(12) + 9(16) = 48 + 144 = 192$.
So,$|\vec{c}| = \sqrt{192} = \sqrt{64 \times 3} = 8\sqrt{3}$.
Finally,$\cos \theta = \frac{-48}{4 \times 8\sqrt{3}} = \frac{-48}{32\sqrt{3}} = \frac{-3}{2\sqrt{3}} = -\frac{\sqrt{3}}{2}$.
Therefore,$\theta = \cos^{-1}\left(-\frac{\sqrt{3}}{2}\right)$.

(B) The area of a parallelogram with diagonals $\overrightarrow{d_1}$ and $\overrightarrow{d_2}$ is given by $\text{Area} = \frac{1}{2} |\overrightarrow{d_1} \times \overrightarrow{d_2}|$.
Here,$\overrightarrow{d_1} = \overrightarrow{AC} = \vec{C} - \vec{A} = (-3-2)\hat{i} + (4-3)\hat{j} + (-2-5)\hat{k} = -5\hat{i} + \hat{j} - 7\hat{k}$.
The second diagonal is $\overrightarrow{d_2} = \overrightarrow{BD} = \hat{i} + 2\hat{j} + 3\hat{k}$.
Now,calculate the cross product $\overrightarrow{AC} \times \overrightarrow{BD}$:
$\overrightarrow{AC} \times \overrightarrow{BD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -5 & 1 & -7 \\ 1 & 2 & 3 \end{vmatrix} = \hat{i}(3 - (-14)) - \hat{j}(-15 - (-7)) + \hat{k}(-10 - 1) = 17\hat{i} + 8\hat{j} - 11\hat{k}$.
The magnitude is $|17\hat{i} + 8\hat{j} - 11\hat{k}| = \sqrt{17^2 + 8^2 + (-11)^2} = \sqrt{289 + 64 + 121} = \sqrt{474}$.
Thus,the area is $\frac{1}{2} \sqrt{474}$.

(B) Given that $|\vec{b}|=1$ and $|\vec{b} \times \vec{a}|=2$.
We know that the cross product $\vec{b} \times \vec{a}$ is a vector perpendicular to both $\vec{b}$ and $\vec{a}$.
Therefore,$(\vec{b} \times \vec{a}) \cdot \vec{b} = 0$.
We need to calculate $|(\vec{b} \times \vec{a}) - \vec{b}|^2$.
Using the property $|\vec{u} - \vec{v}|^2 = |\vec{u}|^2 + |\vec{v}|^2 - 2(\vec{u} \cdot \vec{v})$,we get:
$|(\vec{b} \times \vec{a}) - \vec{b}|^2 = |\vec{b} \times \vec{a}|^2 + |\vec{b}|^2 - 2((\vec{b} \times \vec{a}) \cdot \vec{b})$.
Since $(\vec{b} \times \vec{a}) \cdot \vec{b} = 0$,the expression simplifies to:
$|(\vec{b} \times \vec{a}) - \vec{b}|^2 = |\vec{b} \times \vec{a}|^2 + |\vec{b}|^2$.
Substituting the given values:
$|(\vec{b} \times \vec{a}) - \vec{b}|^2 = (2)^2 + (1)^2 = 4 + 1 = 5$.

(A) Given $L_1$ is perpendicular to $L_2$. The direction vectors are $\vec{v_1} = \hat{i} - \hat{j} + 2\hat{k}$ and $\vec{v_2} = 3\hat{i} + \hat{j} + p\hat{k}$.
Since $L_1 \perp L_2$,their dot product is zero: $(1)(3) + (-1)(1) + (2)(p) = 0$.
$3 - 1 + 2p = 0 \implies 2p = -2 \implies p = -1$.
Line $L_3$ is perpendicular to both $L_1$ and $L_2$,so its direction vector $\vec{v_3}$ is parallel to $\vec{v_1} \times \vec{v_2}$.
$\vec{v_3} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 2 \\ 3 & 1 & -1 \end{vmatrix} = \hat{i}(1 - 2) - \hat{j}(-1 - 6) + \hat{k}(1 + 3) = -\hat{i} + 7\hat{j} + 4\hat{k}$.
Thus,the equation of $L_3$ is $\overrightarrow{r} = \delta(-\hat{i} + 7\hat{j} + 4\hat{k})$.
Any point on $L_3$ is of the form $(-\delta, 7\delta, 4\delta)$.
For $\delta = 1$,the point is $(-1, 7, 4)$.

(D) Given $\vec{c} = 2(\vec{a} \times \vec{b}) - 3\vec{b}$.
Taking the dot product with $\vec{b}$:
$\vec{b} \cdot \vec{c} = \vec{b} \cdot (2(\vec{a} \times \vec{b}) - 3\vec{b}) = 2(\vec{b} \cdot (\vec{a} \times \vec{b})) - 3|\vec{b}|^2$.
Since $\vec{b} \cdot (\vec{a} \times \vec{b}) = 0$,we have $\vec{b} \cdot \vec{c} = -3|\vec{b}|^2 = -3(4)^2 = -48$.
Also,$\vec{b} \cdot \vec{c} = |\vec{b}||\vec{c}| \cos \alpha = 4|\vec{c}| \cos \alpha$.
Thus,$4|\vec{c}| \cos \alpha = -48 \Rightarrow |\vec{c}| \cos \alpha = -12$.
Now,calculate $|\vec{c}|^2$:
$|\vec{c}|^2 = |2(\vec{a} \times \vec{b}) - 3\vec{b}|^2 = 4|\vec{a} \times \vec{b}|^2 + 9|\vec{b}|^2 - 12(\vec{a} \times \vec{b}) \cdot \vec{b}$.
Since $(\vec{a} \times \vec{b}) \cdot \vec{b} = 0$,$|\vec{c}|^2 = 4|\vec{a} \times \vec{b}|^2 + 9(16) = 4|\vec{a}|^2|\vec{b}|^2 \sin^2 \theta + 144$.
Given $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos \theta = 1 \cdot 4 \cdot \cos \theta = 2 \Rightarrow \cos \theta = 1/2 \Rightarrow \sin^2 \theta = 3/4$.
$|\vec{c}|^2 = 4(1)(16)(3/4) + 144 = 48 + 144 = 192$.
We have $|\vec{c}|^2 \cos^2 \alpha = (-12)^2 = 144$.
$192 \cos^2 \alpha = 144 \Rightarrow \cos^2 \alpha = 144/192 = 3/4$.
Then $\sin^2 \alpha = 1 - \cos^2 \alpha = 1 - 3/4 = 1/4$.
Therefore,$192 \sin^2 \alpha = 192 \times (1/4) = 48$.

(B) Given $\overrightarrow{a}=3 \hat{i}+2 \hat{j}+\hat{k}$ and $\overrightarrow{b}=2 \hat{i}-\hat{j}+3 \hat{k}$.
First,calculate $\overrightarrow{a}+\overrightarrow{b} = 5 \hat{i}+\hat{j}+4 \hat{k}$ and $\overrightarrow{a} \times \overrightarrow{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 1 \\ 2 & -1 & 3 \end{vmatrix} = \hat{i}(6+1) - \hat{j}(9-2) + \hat{k}(-3-4) = 7 \hat{i}-7 \hat{j}-7 \hat{k}$.
The equation $(\overrightarrow{a}+\overrightarrow{b}) \times \overrightarrow{c} = 2(\overrightarrow{a} \times \overrightarrow{b}) + 24 \hat{j}-6 \hat{k}$ becomes:
$(5 \hat{i}+\hat{j}+4 \hat{k}) \times \overrightarrow{c} = 2(7 \hat{i}-7 \hat{j}-7 \hat{k}) + 24 \hat{j}-6 \hat{k} = 14 \hat{i}+10 \hat{j}-20 \hat{k}$.
Let $\overrightarrow{c} = x \hat{i}+y \hat{j}+z \hat{k}$. Then $\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 5 & 1 & 4 \\ x & y & z \end{vmatrix} = 14 \hat{i}+10 \hat{j}-20 \hat{k}$.
This gives $\hat{i}(z-4y) - \hat{j}(5z-4x) + \hat{k}(5y-x) = 14 \hat{i}+10 \hat{j}-20 \hat{k}$.
Equating components: $z-4y=14$,$4x-5z=10$,$5y-x=-20$.
Also,$(\overrightarrow{a}-\overrightarrow{b}+\hat{i}) \cdot \overrightarrow{c} = -3$. Since $\overrightarrow{a}-\overrightarrow{b}+\hat{i} = (3-2+1)\hat{i} + (2-(-1))\hat{j} + (1-3)\hat{k} = 2 \hat{i}+3 \hat{j}-2 \hat{k}$,we have $2x+3y-2z=-3$.
Solving the system: From $x=5y+20$,substitute into others to get $x=5, y=-3, z=2$.
Thus,$|\overrightarrow{c}|^2 = 5^2 + (-3)^2 + 2^2 = 25+9+4 = 38$.

(A) The area of triangle $ABC$ is given by $\text{Area} = \frac{1}{2} |\overrightarrow{AB} \times \overrightarrow{AC}| = 15 \sqrt{2}$.
First,calculate the cross product $\overrightarrow{AB} \times \overrightarrow{AC}$:
$\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -7 \\ 6 & d & -2 \end{vmatrix} = \hat{i}(-4 + 7d) - \hat{j}(-2 + 42) + \hat{k}(d - 12) = (7d - 4)\hat{i} - 40\hat{j} + (d - 12)\hat{k}$.
Now,find the magnitude:
$|\overrightarrow{AB} \times \overrightarrow{AC}|^2 = (7d - 4)^2 + (-40)^2 + (d - 12)^2 = (2 \times 15 \sqrt{2})^2 = 900 \times 2 = 1800$.
$(49d^2 - 56d + 16) + 1600 + (d^2 - 24d + 144) = 1800$.
$50d^2 - 80d + 1760 = 1800 \implies 50d^2 - 80d - 40 = 0 \implies 5d^2 - 8d - 4 = 0$.
Solving for $d$:
$5d^2 - 10d + 2d - 4 = 0 \implies 5d(d - 2) + 2(d - 2) = 0 \implies (5d + 2)(d - 2) = 0$.
Since $d > 0$,we have $d = 2$.
Using the vector triangle law $\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}$,we have $\overrightarrow{BC} = \overrightarrow{AC} - \overrightarrow{AB} = (6-1)\hat{i} + (d-2)\hat{j} + (-2 - (-7))\hat{k} = 5\hat{i} + 0\hat{j} + 5\hat{k}$.
Now,calculate the squares of the lengths of the sides:
$|\overrightarrow{AB}|^2 = 1^2 + 2^2 + (-7)^2 = 1 + 4 + 49 = 54$.
$|\overrightarrow{BC}|^2 = 5^2 + 0^2 + 5^2 = 25 + 25 = 50$.
$|\overrightarrow{AC}|^2 = 6^2 + 2^2 + (-2)^2 = 36 + 4 + 4 = 44$.
The largest side is $\sqrt{54}$,and its square is $54$.

(A) The area of a quadrilateral with vertices $A, B, C, D$ is given by the formula $\text{Area} = \frac{1}{2} |\overrightarrow{AC} \times \overrightarrow{BD}|$.
First,we find the vectors $\overrightarrow{AC}$ and $\overrightarrow{BD}$:
$\overrightarrow{AC} = (3-1)\hat{i} + (4-(-1))\hat{j} + (-10-2)\hat{k} = 2\hat{i} + 5\hat{j} - 12\hat{k}$.
$\overrightarrow{BD} = (-1-5)\hat{i} + (-4-7)\hat{j} + (-2-(-6))\hat{k} = -6\hat{i} - 11\hat{j} + 4\hat{k}$.
Now,calculate the cross product $\overrightarrow{AC} \times \overrightarrow{BD}$:
$\overrightarrow{AC} \times \overrightarrow{BD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 5 & -12 \\ -6 & -11 & 4 \end{vmatrix} = \hat{i}(20 - 132) - \hat{j}(8 - 72) + \hat{k}(-22 + 30) = -112\hat{i} + 64\hat{j} + 8\hat{k}$.
The magnitude is $|\overrightarrow{AC} \times \overrightarrow{BD}| = \sqrt{(-112)^2 + 64^2 + 8^2} = \sqrt{12544 + 4096 + 64} = \sqrt{16704} = \sqrt{576 \times 29} = 24\sqrt{29}$.
Finally,the area is $\frac{1}{2} \times 24\sqrt{29} = 12\sqrt{29}$.

(A) The area of a quadrilateral with diagonals $\vec{d_1}$ and $\vec{d_2}$ is given by $\frac{1}{2} |\vec{d_1} \times \vec{d_2}|$.
Here,the diagonals are $\vec{AC}$ and $\vec{BD}$.
$\vec{AC} = \vec{C} - \vec{A} = (2-3)\hat{i} + (2-1)\hat{j} + (1 - (-1))\hat{k} = -\hat{i} + \hat{j} + 2\hat{k}$.
$\vec{BD} = \vec{D} - \vec{B} = \left(\frac{10}{3} - \frac{5}{3}\right)\hat{i} + \left(\frac{2}{3} - \frac{7}{3}\right)\hat{j} + \left(-\frac{1}{3} - \frac{1}{3}\right)\hat{k} = \frac{5}{3}\hat{i} - \frac{5}{3}\hat{j} - \frac{2}{3}\hat{k}$.
Now,calculate the cross product $\vec{AC} \times \vec{BD}$:
$\vec{AC} \times \vec{BD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & 2 \\ \frac{5}{3} & -\frac{5}{3} & -\frac{2}{3} \end{vmatrix} = \hat{i}\left(-\frac{2}{3} - \left(-\frac{10}{3}\right)\right) - \hat{j}\left(\frac{2}{3} - \frac{10}{3}\right) + \hat{k}\left(\frac{5}{3} - \frac{5}{3}\right) = \frac{8}{3}\hat{i} + \frac{8}{3}\hat{j} + 0\hat{k}$.
The magnitude is $|\vec{AC} \times \vec{BD}| = \sqrt{\left(\frac{8}{3}\right)^2 + \left(\frac{8}{3}\right)^2} = \sqrt{\frac{64}{9} + \frac{64}{9}} = \sqrt{\frac{128}{9}} = \frac{8\sqrt{2}}{3}$.
Area $= \frac{1}{2} |\vec{AC} \times \vec{BD}| = \frac{1}{2} \times \frac{8\sqrt{2}}{3} = \frac{4\sqrt{2}}{3}$.

(D) Given $\vec{a} = 6\hat{i} + \hat{j} - \hat{k}$,so $|\vec{a}|^2 = 6^2 + 1^2 + (-1)^2 = 36 + 1 + 1 = 38$.
Given $|\vec{c} - \vec{a}| = 2\sqrt{2}$,squaring both sides gives $|\vec{c}|^2 + |\vec{a}|^2 - 2(\vec{c} \cdot \vec{a}) = 8$.
Substituting $\vec{a} \cdot \vec{c} = 6|\vec{c}|$ and $|\vec{a}|^2 = 38$,we get $|\vec{c}|^2 + 38 - 2(6|\vec{c}|) = 8$.
$|\vec{c}|^2 - 12|\vec{c}| + 30 = 0$.
Solving for $|\vec{c}|$ using the quadratic formula: $|\vec{c}| = \frac{12 \pm \sqrt{144 - 120}}{2} = \frac{12 \pm \sqrt{24}}{2} = 6 \pm \sqrt{6}$.
Since $|\vec{c}| \geq 6$,we take $|\vec{c}| = 6 + \sqrt{6}$.
Now,$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 6 & 1 & -1 \\ 1 & 1 & 0 \end{vmatrix} = \hat{i}(0 - (-1)) - \hat{j}(0 - (-1)) + \hat{k}(6 - 1) = \hat{i} - \hat{j} + 5\hat{k}$.
$|\vec{a} \times \vec{b}| = \sqrt{1^2 + (-1)^2 + 5^2} = \sqrt{1 + 1 + 25} = \sqrt{27} = 3\sqrt{3}$.
The magnitude of the cross product is $|(\vec{a} \times \vec{b}) \times \vec{c}| = |\vec{a} \times \vec{b}| |\vec{c}| \sin(60^{\circ})$.
$|(\vec{a} \times \vec{b}) \times \vec{c}| = (3\sqrt{3}) (6 + \sqrt{6}) \frac{\sqrt{3}}{2} = \frac{3 \times 3}{2} (6 + \sqrt{6}) = \frac{9}{2}(6 + \sqrt{6})$.

(B) Given $\vec{a} = 2\hat{i} + \hat{j} - \hat{k}$.
First,calculate $\vec{v} = \vec{a} \times (\hat{i} + \hat{j})$:
$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ 1 & 1 & 0 \end{vmatrix} = \hat{i}(0 - (-1)) - \hat{j}(0 - (-1)) + \hat{k}(2 - 1) = \hat{i} - \hat{j} + \hat{k}$.
Next,calculate $\vec{w} = \vec{v} \times \hat{i} = (\hat{i} - \hat{j} + \hat{k}) \times \hat{i} = \hat{i} \times \hat{i} - \hat{j} \times \hat{i} + \hat{k} \times \hat{i} = 0 - (-\hat{k}) + \hat{j} = \hat{j} + \hat{k}$.
Then,$\vec{b} = \vec{w} \times \hat{i} = (\hat{j} + \hat{k}) \times \hat{i} = \hat{j} \times \hat{i} + \hat{k} \times \hat{i} = -\hat{k} + \hat{j} = \hat{j} - \hat{k}$.
The projection of $\vec{a}$ on $\vec{b}$ is given by $p = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$.
$\vec{a} \cdot \vec{b} = (2\hat{i} + \hat{j} - \hat{k}) \cdot (\hat{j} - \hat{k}) = 0 + 1 + 1 = 2$.
$|\vec{b}| = \sqrt{0^2 + 1^2 + (-1)^2} = \sqrt{2}$.
So,$p = \frac{2}{\sqrt{2}} = \sqrt{2}$.
The square of the projection is $p^2 = (\sqrt{2})^2 = 2$.

(D) The area of a parallelogram with adjacent sides $\overrightarrow{OA}$ and $\overrightarrow{OC}$ is given by $|\overrightarrow{OA} \times \overrightarrow{OC}|$.
Given,$|\overrightarrow{OA} \times \overrightarrow{OC}| = |2 \overrightarrow{a} \times 3 \overrightarrow{b}| = 15$.
$6 |\overrightarrow{a} \times \overrightarrow{b}| = 15 \implies |\overrightarrow{a} \times \overrightarrow{b}| = \frac{15}{6} = \frac{5}{2} \dots (1)$.
The area of a quadrilateral $OABC$ with diagonals $\overrightarrow{OB}$ and $\overrightarrow{AC}$ is given by $\frac{1}{2} |\overrightarrow{OB} \times \overrightarrow{AC}|$.
We have $\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = 3 \overrightarrow{b} - 2 \overrightarrow{a}$.
Area $= \frac{1}{2} |(6 \overrightarrow{a} + 5 \overrightarrow{b}) \times (3 \overrightarrow{b} - 2 \overrightarrow{a})|$.
$= \frac{1}{2} |18 (\overrightarrow{a} \times \overrightarrow{b}) - 12 (\overrightarrow{a} \times \overrightarrow{a}) + 15 (\overrightarrow{b} \times \overrightarrow{b}) - 10 (\overrightarrow{b} \times \overrightarrow{a})|$.
Since $\overrightarrow{a} \times \overrightarrow{a} = 0$,$\overrightarrow{b} \times \overrightarrow{b} = 0$,and $\overrightarrow{b} \times \overrightarrow{a} = -(\overrightarrow{a} \times \overrightarrow{b})$:
Area $= \frac{1}{2} |18 (\overrightarrow{a} \times \overrightarrow{b}) + 10 (\overrightarrow{a} \times \overrightarrow{b})| = \frac{1}{2} |28 (\overrightarrow{a} \times \overrightarrow{b})| = 14 |\overrightarrow{a} \times \overrightarrow{b}|$.
Substituting from $(1)$:
Area $= 14 \times \frac{5}{2} = 35$ sq. units.

(D) The area of a parallelogram with diagonals $\vec{d}_1$ and $\vec{d}_2$ is given by $\text{Area} = \frac{1}{2} |\vec{d}_1 \times \vec{d}_2|$.
Given diagonals are $\vec{d}_1 = \vec{a}+\vec{b} = (2-1)\hat{i} + \alpha\hat{j} + (1+1)\hat{k} = \hat{i} + \alpha\hat{j} + 2\hat{k}$ and $\vec{d}_2 = \vec{b}+\vec{c} = -\hat{i} + \beta\hat{j} + (1-1)\hat{k} = -\hat{i} + \beta\hat{j}$.
Calculating the cross product $\vec{d}_1 \times \vec{d}_2$:
$\vec{d}_1 \times \vec{d}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & \alpha & 2 \\ -1 & \beta & 0 \end{vmatrix} = \hat{i}(0 - 2\beta) - \hat{j}(0 - (-2)) + \hat{k}(\beta - (-\alpha)) = -2\beta\hat{i} - 2\hat{j} + (\alpha+\beta)\hat{k}$.
The area is $\frac{1}{2} \sqrt{(-2\beta)^2 + (-2)^2 + (\alpha+\beta)^2} = \frac{\sqrt{21}}{2}$.
Squaring both sides: $4\beta^2 + 4 + \alpha^2 + \beta^2 + 2\alpha\beta = 21$.
Given $\alpha\beta = -6$,substitute this into the equation: $\alpha^2 + 5\beta^2 + 2(-6) + 4 = 21 \implies \alpha^2 + 5\beta^2 - 8 = 21 \implies \alpha^2 + 5\beta^2 = 29$.
Since $\alpha, \beta$ are integers and $\alpha\beta = -6$,we test pairs: If $\beta=2, \alpha=-3$,then $(-3)^2 + 5(2)^2 = 9 + 20 = 29$ (Valid). If $\beta=-2, \alpha=3$,then $(3)^2 + 5(-2)^2 = 9 + 20 = 29$ (Valid).
Let $(\alpha_1, \beta_1) = (-3, 2)$ and $(\alpha_2, \beta_2) = (3, -2)$.
Then $\alpha_1^2 + \beta_1^2 - \alpha_2\beta_2 = (-3)^2 + (2)^2 - (3)(-2) = 9 + 4 + 6 = 19$.

(C) In a regular hexagon $PQRSTU$,the sides are vectors. Let $\vec{a} = \overline{PQ}$. Then $\overline{RS}$ is parallel to $\overline{UP}$ and $\overline{ST}$ is parallel to $\overline{PQ}$.
Specifically,$\overline{RS} = -\overline{PQ} = -\vec{a}$ is incorrect; rather,$\overline{RS}$ is parallel to $\overline{PQ}$ but in the opposite direction,so $\overline{RS} = -\overline{PQ}$ is false. In a regular hexagon,$\overline{PQ} = \overline{UT}$ and $\overline{RS} = \overline{PQ}$ is not true. Actually,$\overline{PQ} \parallel \overline{ST}$ and $\overline{RS} \parallel \overline{UP}$.
Since $\overline{PQ}$ is not parallel to $\overline{RS}$,$\overline{PQ} \times \overline{RS} \neq \overrightarrow{0}$.
Thus,$Statement-2$ is False because it claims $\overline{PQ} \times \overline{RS} = \overrightarrow{0}$.
For $Statement-1$,$\overline{PQ} \times (\overline{RS} + \overline{ST}) = \overline{PQ} \times \overline{RS} + \overline{PQ} \times \overline{ST}$. Since $\overline{PQ}$ is not parallel to the resultant $\overline{RS} + \overline{ST}$,the cross product is non-zero. Thus,$Statement-1$ is True.

(B) Given that $\vec{a}, \vec{b}, \vec{c}$ are unit vectors such that $\vec{a}+\vec{b}+\vec{c}=\vec{0}$.
Since $\vec{a}+\vec{b}+\vec{c}=\vec{0}$,we have $\vec{a}+\vec{b}=-\vec{c}$.
Taking the cross product with $\vec{a}$ on both sides: $\vec{a} \times (\vec{a}+\vec{b}) = \vec{a} \times (-\vec{c})$.
This gives $\vec{a} \times \vec{a} + \vec{a} \times \vec{b} = -\vec{a} \times \vec{c}$,which simplifies to $\vec{0} + \vec{a} \times \vec{b} = \vec{c} \times \vec{a}$.
So,$\vec{a} \times \vec{b} = \vec{c} \times \vec{a}$.
Similarly,taking the cross product with $\vec{b}$ on both sides: $\vec{b} \times (\vec{a}+\vec{b}) = \vec{b} \times (-\vec{c})$.
This gives $\vec{b} \times \vec{a} + \vec{b} \times \vec{b} = -\vec{b} \times \vec{c}$,which simplifies to $\vec{b} \times \vec{a} = \vec{c} \times \vec{b}$,or $\vec{a} \times \vec{b} = \vec{b} \times \vec{c}$.
Thus,$\vec{a} \times \vec{b} = \vec{b} \times \vec{c} = \vec{c} \times \vec{a}$.
Since $\vec{a}, \vec{b}, \vec{c}$ are unit vectors forming an equilateral triangle,the cross products are non-zero. Therefore,$\vec{a} \times \vec{b} = \vec{b} \times \vec{c} = \vec{c} \times \vec{a} \neq \vec{0}$.

(D) Given $\hat{u} \times \vec{w} = \hat{v}$. Since $\hat{u}$ and $\vec{w}$ are unit vectors,$|\hat{u} \times \vec{w}| = |\hat{u}| |\vec{w}| \sin \theta = \sin \theta$. Also,$|\hat{v}| = \frac{1}{\sqrt{6}} \sqrt{1^2 + 1^2 + 2^2} = 1$. Thus,$\sin \theta = 1$,which implies $\theta = 90^{\circ}$.
Since $\hat{u} \times \vec{w} = \hat{v}$,$\hat{v}$ is perpendicular to both $\hat{u}$ and $\vec{w}$.
For a fixed $\hat{u}$,there are infinitely many vectors $\vec{w}$ perpendicular to $\hat{u}$ such that their cross product with $\hat{u}$ is $\hat{v}$. Thus,option $B$ is correct.
Since $\hat{v} \cdot \hat{u} = 0$,we have $\frac{1}{\sqrt{6}}(u_1 + u_2 + 2u_3) = 0$,so $u_1 + u_2 + 2u_3 = 0$.
If $\hat{u}$ lies in the $xy$-plane,$u_3 = 0$,then $u_1 + u_2 = 0$,so $|u_1| = |u_2|$. Thus,option $C$ is correct.
If $\hat{u}$ lies in the $xz$-plane,$u_2 = 0$,then $u_1 + 2u_3 = 0$,so $|u_1| = 2|u_3|$. Thus,option $D$ is incorrect as it states $2|u_1| = |u_3|$.

(D) $(1)$ Since $\overline{OX}, \overline{OY}$ are unit vectors along $QR$ and $RP$,the angle between them is $\pi - R$.
Thus,$|\overline{OX} \times \overline{OY}| = |\overline{OX}| |\overline{OY}| \sin(\pi - R) = 1 \cdot 1 \cdot \sin R = \sin R$.
Since $P+Q+R = \pi$,$\sin R = \sin(\pi - (P+Q)) = \sin(P+Q)$.
Thus,option $A$ is correct.
$(2)$ We need to find the minimum value of $\cos(P+Q) + \cos(Q+R) + \cos(R+P)$.
Since $P+Q+R = \pi$,this is equivalent to $\cos(\pi-R) + \cos(\pi-P) + \cos(\pi-Q) = -(\cos P + \cos Q + \cos R)$.
For any triangle,$\cos P + \cos Q + \cos R \leq \frac{3}{2}$.
Therefore,$-(\cos P + \cos Q + \cos R) \geq -\frac{3}{2}$.
The minimum value is $-\frac{3}{2}$,which occurs for an equilateral triangle.
Thus,option $B$ is correct.
Combining both,the correct choice is $A, B$.

(C) The vertices of the cube are $O(0,0,0), P(1,0,0), Q(0,1,0), R(0,0,1)$ and $T(1,1,1)$. The center $S$ is $\left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right)$.
Calculating the vectors:
$\overrightarrow{p} = \overrightarrow{SP} = (1-\frac{1}{2})\hat{i} + (0-\frac{1}{2})\hat{j} + (0-\frac{1}{2})\hat{k} = \frac{1}{2}\hat{i} - \frac{1}{2}\hat{j} - \frac{1}{2}\hat{k}$
$\overrightarrow{q} = \overrightarrow{SQ} = (0-\frac{1}{2})\hat{i} + (1-\frac{1}{2})\hat{j} + (0-\frac{1}{2})\hat{k} = -\frac{1}{2}\hat{i} + \frac{1}{2}\hat{j} - \frac{1}{2}\hat{k}$
$\overrightarrow{r} = \overrightarrow{SR} = (0-\frac{1}{2})\hat{i} + (0-\frac{1}{2})\hat{j} + (1-\frac{1}{2})\hat{k} = -\frac{1}{2}\hat{i} - \frac{1}{2}\hat{j} + \frac{1}{2}\hat{k}$
$\overrightarrow{t} = \overrightarrow{ST} = (1-\frac{1}{2})\hat{i} + (1-\frac{1}{2})\hat{j} + (1-\frac{1}{2})\hat{k} = \frac{1}{2}\hat{i} + \frac{1}{2}\hat{j} + \frac{1}{2}\hat{k}$
Calculating cross products:
$\overrightarrow{p} \times \overrightarrow{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1/2 & -1/2 & -1/2 \\ -1/2 & 1/2 & -1/2 \end{vmatrix} = \hat{i}(\frac{1}{4} + \frac{1}{4}) - \hat{j}(-\frac{1}{4} - \frac{1}{4}) + \hat{k}(\frac{1}{4} - \frac{1}{4}) = \frac{1}{2}\hat{i} + \frac{1}{2}\hat{j}$
$\overrightarrow{r} \times \overrightarrow{t} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1/2 & -1/2 & 1/2 \\ 1/2 & 1/2 & 1/2 \end{vmatrix} = \hat{i}(-\frac{1}{4} - \frac{1}{4}) - \hat{j}(-\frac{1}{4} - \frac{1}{4}) + \hat{k}(-\frac{1}{4} + \frac{1}{4}) = -\frac{1}{2}\hat{i} + \frac{1}{2}\hat{j}$
Now,$|(\overrightarrow{p} \times \overrightarrow{q}) \times (\overrightarrow{r} \times \overrightarrow{t})| = |(\frac{1}{2}\hat{i} + \frac{1}{2}\hat{j}) \times (-\frac{1}{2}\hat{i} + \frac{1}{2}\hat{j})| = |\frac{1}{4}(\hat{i} + \hat{j}) \times (-\hat{i} + \hat{j})| = |\frac{1}{4}(\hat{k} - (-\hat{k}))| = |\frac{1}{4}(2\hat{k})| = |\frac{1}{2}\hat{k}| = \frac{1}{2} = 0.5$

(C) Given $\overrightarrow{a}=3\hat{i}-\hat{j}+2\hat{k}$.
First,calculate $\overrightarrow{b}=\overrightarrow{a}\times(\hat{i}-2\hat{k})$:
$\overrightarrow{b}=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 2 \\ 1 & 0 & -2\end{vmatrix} = \hat{i}(2-0) - \hat{j}(-6-2) + \hat{k}(0-(-1)) = 2\hat{i}+8\hat{j}+\hat{k}$.
Next,calculate $\overrightarrow{c}=\overrightarrow{b}\times\hat{k}$:
$\overrightarrow{c}=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 2 & 8 & 1 \\ 0 & 0 & 1\end{vmatrix} = \hat{i}(8-0) - \hat{j}(2-0) + \hat{k}(0-0) = 8\hat{i}-2\hat{j}$.
Now,find $\overrightarrow{c}-2\hat{j} = (8\hat{i}-2\hat{j}) - 2\hat{j} = 8\hat{i}-4\hat{j}$.
The projection of a vector $\overrightarrow{v}$ on $\overrightarrow{a}$ is given by $\frac{\overrightarrow{v} \cdot \overrightarrow{a}}{|\overrightarrow{a}|}$.
Here,$\overrightarrow{v} = 8\hat{i}-4\hat{j}$ and $\overrightarrow{a} = 3\hat{i}-\hat{j}+2\hat{k}$.
$|\overrightarrow{a}| = \sqrt{3^2+(-1)^2+2^2} = \sqrt{9+1+4} = \sqrt{14}$.
Projection $= \frac{(8\hat{i}-4\hat{j}) \cdot (3\hat{i}-\hat{j}+2\hat{k})}{\sqrt{14}} = \frac{(8)(3) + (-4)(-1) + (0)(2)}{\sqrt{14}} = \frac{24+4}{\sqrt{14}} = \frac{28}{\sqrt{14}} = 2\sqrt{14}$.

(C) Given $\overrightarrow{a}=2 \hat{i}-\hat{j}+3 \hat{k}$ and $\overrightarrow{b}=3 \hat{i}-5 \hat{j}+\hat{k}$.
From $\overrightarrow{a} \times \overrightarrow{c}=\overrightarrow{c} \times \overrightarrow{b}$,we have $\overrightarrow{a} \times \overrightarrow{c}+\overrightarrow{b} \times \overrightarrow{c}=0$,which implies $(\overrightarrow{a}+\overrightarrow{b}) \times \overrightarrow{c}=0$.
This means $\overrightarrow{c}$ is parallel to $(\overrightarrow{a}+\overrightarrow{b})$.
Let $\overrightarrow{c}=\lambda(\overrightarrow{a}+\overrightarrow{b})=\lambda(5 \hat{i}-6 \hat{j}+4 \hat{k})$.
Then $|\overrightarrow{c}|^2=\lambda^2(5^2+(-6)^2+4^2)=77 \lambda^2$.
Given $(\overrightarrow{a}+\overrightarrow{c}) \cdot(\overrightarrow{b}+\overrightarrow{c})=168$.
Expanding this,we get $\overrightarrow{a} \cdot \overrightarrow{b}+\overrightarrow{a} \cdot \overrightarrow{c}+\overrightarrow{c} \cdot \overrightarrow{b}+|\overrightarrow{c}|^2=168$.
Calculating $\overrightarrow{a} \cdot \overrightarrow{b} = (2)(3)+(-1)(-5)+(3)(1) = 6+5+3=14$.
Substituting $\overrightarrow{c}=\lambda(\overrightarrow{a}+\overrightarrow{b})$,we get $14+\lambda(\overrightarrow{a}+\overrightarrow{b}) \cdot (\overrightarrow{a}+\overrightarrow{b})+77 \lambda^2=168$.
$14+\lambda|\overrightarrow{a}+\overrightarrow{b}|^2+77 \lambda^2=168$.
$14+\lambda(77)+77 \lambda^2=168$.
$77 \lambda^2+77 \lambda-154=0 \Rightarrow \lambda^2+\lambda-2=0$.
Solving for $\lambda$,we get $(\lambda+2)(\lambda-1)=0$,so $\lambda=-2$ or $\lambda=1$.
Since $|\overrightarrow{c}|^2=77 \lambda^2$,the maximum value occurs at $\lambda=-2$.
$|\overrightarrow{c}|^2=77 \times (-2)^2 = 77 \times 4 = 308$.

(D) Let $\overrightarrow{v} = \hat{i} + \hat{j} + \hat{k}$.
First,find the cross product $\overrightarrow{b} \times \overrightarrow{c}$:
$\overrightarrow{b} \times \overrightarrow{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 2 & 3 & -1 \end{vmatrix} = \hat{i}(2 - 9) - \hat{j}(-1 - 6) + \hat{k}(3 + 4) = -7\hat{i} + 7\hat{j} + 7\hat{k} = -7(\hat{i} - \hat{j} - \hat{k})$.
Since $\hat{a}$ is a unit vector perpendicular to $\overrightarrow{b}$ and $\overrightarrow{c}$,$\hat{a} = \pm \frac{\hat{i} - \hat{j} - \hat{k}}{\sqrt{3}}$.
Given $\hat{a} \cdot \frac{\overrightarrow{v}}{|\overrightarrow{v}|} = \cos\left(\cos^{-1}\left(-\frac{1}{3}\right)\right) = -\frac{1}{3}$.
If $\hat{a} = \frac{\hat{i} - \hat{j} - \hat{k}}{\sqrt{3}}$,then $\hat{a} \cdot \frac{\overrightarrow{v}}{\sqrt{3}} = \frac{1 - 1 - 1}{3} = -\frac{1}{3}$. This satisfies the condition.
If $\hat{a} = \frac{-\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}$,then $\hat{a} \cdot \frac{\overrightarrow{v}}{\sqrt{3}} = \frac{-1 + 1 + 1}{3} = \frac{1}{3}$. This is rejected.
So,$\hat{a} = \frac{\hat{i} - \hat{j} - \hat{k}}{\sqrt{3}}$.
Now,$\hat{a}$ makes an angle $\frac{\pi}{3}$ with $\overrightarrow{u} = \hat{i} + \alpha\hat{j} + \hat{k}$.
$\cos\left(\frac{\pi}{3}\right) = \frac{\hat{a} \cdot \overrightarrow{u}}{|\hat{a}| |\overrightarrow{u}|} \Rightarrow \frac{1}{2} = \frac{\frac{1 - \alpha - 1}{\sqrt{3}}}{\sqrt{1 + \alpha^2 + 1}} = \frac{-\alpha}{\sqrt{3}\sqrt{\alpha^2 + 2}}$.
$\frac{\sqrt{3}}{2} \sqrt{\alpha^2 + 2} = -\alpha$. Since $\alpha < 0$,squaring both sides gives $\frac{3}{4}(\alpha^2 + 2) = \alpha^2$.
$3\alpha^2 + 6 = 4\alpha^2 \Rightarrow \alpha^2 = 6$. Since $\alpha < 0$,$\alpha = -\sqrt{6}$.

(D) Given $\vec{a}=2 \hat{i}-3 \hat{j}+\hat{k}$ and $\vec{b}=3 \hat{i}+2 \hat{j}+5 \hat{k}$.
First,calculate the cross product $\vec{a} \times \vec{b}$:
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & 1 \\ 3 & 2 & 5 \end{vmatrix} = \hat{i}(-15-2) - \hat{j}(10-3) + \hat{k}(4+9) = -17 \hat{i}-7 \hat{j}+13 \hat{k}$.
We are given $(\vec{a}-\vec{c}) \times \vec{b} = -18 \hat{i}-3 \hat{j}+12 \hat{k}$.
Using the distributive property of the cross product,we have $(\vec{a} \times \vec{b}) - (\vec{c} \times \vec{b}) = -18 \hat{i}-3 \hat{j}+12 \hat{k}$.
Since $\vec{c} \times \vec{b} = -(\vec{b} \times \vec{c})$,we have $(\vec{a} \times \vec{b}) + (\vec{b} \times \vec{c}) = -18 \hat{i}-3 \hat{j}+12 \hat{k}$.
Substituting $\vec{a} \times \vec{b}$,we get $(-17 \hat{i}-7 \hat{j}+13 \hat{k}) + (\vec{b} \times \vec{c}) = -18 \hat{i}-3 \hat{j}+12 \hat{k}$.
Thus,$\vec{d} = \vec{b} \times \vec{c} = (-18 \hat{i}-3 \hat{j}+12 \hat{k}) - (-17 \hat{i}-7 \hat{j}+13 \hat{k}) = -\hat{i}+4 \hat{j}-\hat{k}$.
Finally,calculate $\vec{a} \cdot \vec{d} = (2 \hat{i}-3 \hat{j}+\hat{k}) \cdot (-\hat{i}+4 \hat{j}-\hat{k}) = (2)(-1) + (-3)(4) + (1)(-1) = -2 - 12 - 1 = -15$.
Therefore,$|\vec{a} \cdot \vec{d}| = |-15| = 15$.

(D) Given $\overrightarrow{b} \times \overrightarrow{d} = \overrightarrow{c} \times \overrightarrow{d}$,we have $(\overrightarrow{b} - \overrightarrow{c}) \times \overrightarrow{d} = \overrightarrow{0}$.
This implies $\overrightarrow{d} = \lambda(\overrightarrow{b} - \overrightarrow{c})$ for some scalar $\lambda$.
Calculating $\overrightarrow{b} - \overrightarrow{c} = (3-2)\hat{i} + (-3 - (-1))\hat{j} + (3-2)\hat{k} = \hat{i} - 2\hat{j} + \hat{k}$.
So,$\overrightarrow{d} = \lambda(\hat{i} - 2\hat{j} + \hat{k})$.
Given $\overrightarrow{a} \cdot \overrightarrow{d} = 4$,we have $(\hat{i} + 2\hat{j} + \hat{k}) \cdot \lambda(\hat{i} - 2\hat{j} + \hat{k}) = 4$.
$\lambda(1 - 4 + 1) = 4 \Rightarrow -2\lambda = 4 \Rightarrow \lambda = -2$.
Thus,$\overrightarrow{d} = -2(\hat{i} - 2\hat{j} + \hat{k}) = -2\hat{i} + 4\hat{j} - 2\hat{k}$.
Now,$\overrightarrow{a} \times \overrightarrow{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 1 \\ -2 & 4 & -2 \end{vmatrix} = \hat{i}(-4 - 4) - \hat{j}(-2 - (-2)) + \hat{k}(4 - (-4)) = -8\hat{i} + 8\hat{k}$.
Then $|\overrightarrow{a} \times \overrightarrow{d}|^2 = (-8)^2 + 0^2 + 8^2 = 64 + 64 = 128$.

(B) Let the vertices be $A(1,2,0)$,$B(1,0,2)$,and $C(0,3,1)$.
The vectors representing the sides are $\vec{AB} = (1-1)\hat{i} + (0-2)\hat{j} + (2-0)\hat{k} = -2\hat{j} + 2\hat{k}$ and $\vec{AC} = (0-1)\hat{i} + (3-2)\hat{j} + (1-0)\hat{k} = -\hat{i} + \hat{j} + \hat{k}$.
The area of $\triangle ABC$ is given by $\frac{1}{2} |\vec{AB} \times \vec{AC}|$.
Calculating the cross product: $\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & -2 & 2 \\ -1 & 1 & 1 \end{vmatrix} = \hat{i}(-2-2) - \hat{j}(0 - (-2)) + \hat{k}(0-2) = -4\hat{i} - 2\hat{j} - 2\hat{k}$.
The magnitude is $|\vec{AB} \times \vec{AC}| = \sqrt{(-4)^2 + (-2)^2 + (-2)^2} = \sqrt{16 + 4 + 4} = \sqrt{24} = 2\sqrt{6}$.
Therefore,the area is $\frac{1}{2} \times 2\sqrt{6} = \sqrt{6}$ sq. units.

(C) Let $A = (1, 2, 0)$,$B = (1, 0, 2)$,and $C = (0, x, 1)$.
$\vec{AB} = (1-1)\hat{i} + (0-2)\hat{j} + (2-0)\hat{k} = -2\hat{j} + 2\hat{k}$.
$\vec{AC} = (0-1)\hat{i} + (x-2)\hat{j} + (1-0)\hat{k} = -\hat{i} + (x-2)\hat{j} + \hat{k}$.
The area of $\triangle ABC$ is given by $\frac{1}{2} |\vec{AB} \times \vec{AC}| = \sqrt{6}$.
$\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & -2 & 2 \\ -1 & x-2 & 1 \end{vmatrix} = \hat{i}(-2 - 2(x-2)) - \hat{j}(0 - (-2)) + \hat{k}(0 - 2) = (2-2x)\hat{i} - 2\hat{j} - 2\hat{k}$.
$|\vec{AB} \times \vec{AC}| = \sqrt{(2-2x)^2 + (-2)^2 + (-2)^2} = \sqrt{4(1-x)^2 + 4 + 4} = \sqrt{4(1-2x+x^2) + 8} = \sqrt{4x^2 - 8x + 12} = 2\sqrt{x^2 - 2x + 3}$.
Given $\frac{1}{2} |\vec{AB} \times \vec{AC}| = \sqrt{6}$,so $\sqrt{x^2 - 2x + 3} = \sqrt{6}$.
Squaring both sides: $x^2 - 2x + 3 = 6 \Rightarrow x^2 - 2x - 3 = 0$.
$(x-3)(x+1) = 0$,so $x = 3$ or $x = -1$.

(C) Let the points be $A(1, -1, 2)$,$B(2, 0, -1)$,and $C(0, 2, 1)$.
First,find two vectors in the plane: $\vec{AB} = (2-1)\hat{i} + (0-(-1))\hat{j} + (-1-2)\hat{k} = \hat{i} + \hat{j} - 3\hat{k}$ and $\vec{AC} = (0-1)\hat{i} + (2-(-1))\hat{j} + (1-2)\hat{k} = -\hat{i} + 3\hat{j} - \hat{k}$.
$A$ vector perpendicular to the plane is given by the cross product $\vec{n} = \vec{AB} \times \vec{AC}$.
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -3 \\ -1 & 3 & -1 \end{vmatrix} = \hat{i}(-1 - (-9)) - \hat{j}(-1 - 3) + \hat{k}(3 - (-1)) = 8\hat{i} + 4\hat{j} + 4\hat{k}$.
Simplifying,we can take $\vec{n}' = 2\hat{i} + \hat{j} + \hat{k}$.
The magnitude is $|\vec{n}'| = \sqrt{2^2 + 1^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$.
The unit vectors are $\pm \frac{\vec{n}'}{|\vec{n}'|} = \pm \frac{2\hat{i} + \hat{j} + \hat{k}}{\sqrt{6}}$.
Thus,the correct option is $C$.

(C) The projection of $\overline{a}$ on $\overline{c}$ is given by $\frac{\overline{a} \cdot \overline{c}}{|\overline{c}|} = \frac{10}{3}$.
$|\overline{c}| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{1+4+4} = 3$.
So,$\frac{(\alpha \hat{i} + 3 \hat{j} - \hat{k}) \cdot (\hat{i} + 2 \hat{j} - 2 \hat{k})}{3} = \frac{10}{3}$.
$\alpha + 6 + 2 = 10 \Rightarrow \alpha = 2$.
Now,$\overline{b} \times \overline{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -\beta & 4 \\ 1 & 2 & -2 \end{vmatrix} = \hat{i}(2\beta - 8) - \hat{j}(-6 - 4) + \hat{k}(6 + \beta) = (2\beta - 8)\hat{i} + 10\hat{j} + (6 + \beta)\hat{k}$.
Given $\overline{b} \times \overline{c} = -6\hat{i} + 10\hat{j} + 7\hat{k}$,comparing components:
$6 + \beta = 7 \Rightarrow \beta = 1$.
Thus,$\alpha^2 + \beta^2 - \alpha\beta = 2^2 + 1^2 - (2)(1) = 4 + 1 - 2 = 3$.

(A) Let $\vec{a} = 2\hat{i}+3\hat{j}+\hat{k}$,$\vec{b} = \hat{i}+\hat{j}+\hat{k}$,and $\vec{c} = \hat{i}+2\hat{j}+3\hat{k}$.
The vector perpendicular to the plane containing $\vec{b}$ and $\vec{c}$ is given by $\vec{n} = \vec{b} \times \vec{c}$.
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & 2 & 3 \end{vmatrix} = \hat{i}(3-2) - \hat{j}(3-1) + \hat{k}(2-1) = \hat{i} - 2\hat{j} + \hat{k}$.
The projection of $\vec{a}$ on $\vec{n}$ is given by $\frac{|\vec{a} \cdot \vec{n}|}{|\vec{n}|}$.
$\vec{a} \cdot \vec{n} = (2\hat{i}+3\hat{j}+\hat{k}) \cdot (\hat{i}-2\hat{j}+\hat{k}) = (2)(1) + (3)(-2) + (1)(1) = 2 - 6 + 1 = -3$.
$|\vec{n}| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{1+4+1} = \sqrt{6}$.
Magnitude of projection = $\frac{|-3|}{\sqrt{6}} = \frac{3}{\sqrt{6}} = \frac{3}{\sqrt{2} \times \sqrt{3}} = \frac{\sqrt{3}}{\sqrt{2}} = \sqrt{\frac{3}{2}}$ units.

(B) Let $\vec{a} = 2\hat{i}+\hat{j}+\hat{k}$,$\vec{b} = \hat{i}+\hat{j}+\hat{k}$,and $\vec{c} = \hat{i}+2\hat{j}+3\hat{k}$.
The vector perpendicular to the plane containing $\vec{b}$ and $\vec{c}$ is given by $\vec{n} = \vec{b} \times \vec{c}$.
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & 2 & 3 \end{vmatrix} = \hat{i}(3-2) - \hat{j}(3-1) + \hat{k}(2-1) = \hat{i} - 2\hat{j} + \hat{k}$.
The magnitude of the projection of $\vec{a}$ on $\vec{n}$ is given by $\left| \frac{\vec{a} \cdot \vec{n}}{|\vec{n}|} \right|$.
$\vec{a} \cdot \vec{n} = (2)(1) + (1)(-2) + (1)(1) = 2 - 2 + 1 = 1$.
$|\vec{n}| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{1 + 4 + 1} = \sqrt{6}$.
Therefore,the required projection is $\frac{1}{\sqrt{6}}$ units.

(B) Let $\vec{a} = 3\hat{i} - 2\hat{j} + 4\hat{k}$ and $\vec{b} = 1\hat{i} + 3\hat{j} - 2\hat{k}$.
Since the line is perpendicular to both lines,its direction vector is given by the cross product $\vec{n} = \vec{a} \times \vec{b}$.
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -2 & 4 \\ 1 & 3 & -2 \end{vmatrix} = \hat{i}(4 - 12) - \hat{j}(-6 - 4) + \hat{k}(9 + 2) = -8\hat{i} + 10\hat{j} + 11\hat{k}$.
The magnitude of the vector is $|\vec{n}| = \sqrt{(-8)^2 + 10^2 + 11^2} = \sqrt{64 + 100 + 121} = \sqrt{285}$.
The direction cosines are the components of the unit vector,which are $\frac{-8}{\sqrt{285}}, \frac{10}{\sqrt{285}}, \frac{11}{\sqrt{285}}$.

(B) Let the direction ratios of the required line be $a, b, c$.
Since the line is perpendicular to the lines with direction ratios $(-1, 2, 2)$ and $(0, 2, 1)$,we have the following equations:
$-a + 2b + 2c = 0$ $(i)$
$0a + 2b + c = 0$ (ii)
From (ii),we get $c = -2b$.
Substituting $c = -2b$ into $(i)$:
$-a + 2b + 2(-2b) = 0$
$-a + 2b - 4b = 0$
$-a - 2b = 0 \implies a = -2b$.
Now,we have $a = -2b$ and $c = -2b$.
If we let $b = 1$,then $a = -2$ and $c = -2$.
The direction ratios are $(-2, 1, -2)$,which is proportional to $(2, -1, 2)$.
Alternatively,using the cross product of the two direction vectors $\vec{n_1} = -\hat{i} + 2\hat{j} + 2\hat{k}$ and $\vec{n_2} = 0\hat{i} + 2\hat{j} + 1\hat{k}$:
$\vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 2 & 2 \\ 0 & 2 & 1 \end{vmatrix} = \hat{i}(2 - 4) - \hat{j}(-1 - 0) + \hat{k}(-2 - 0) = -2\hat{i} + 1\hat{j} - 2\hat{k}$.
The direction ratios are proportional to $(-2, 1, -2)$,which is equivalent to $(2, -1, 2)$.

(B) The direction vectors of the lines $L_1$ and $L_2$ are $\vec{v_1} = 3\hat{i} + \hat{j} + 2\hat{k}$ and $\vec{v_2} = \hat{i} + 2\hat{j} + 3\hat{k}$ respectively.
$A$ vector perpendicular to both $L_1$ and $L_2$ is given by the cross product $\vec{n} = \vec{v_1} \times \vec{v_2}$.
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 2 \\ 1 & 2 & 3 \end{vmatrix} = \hat{i}(3-4) - \hat{j}(9-2) + \hat{k}(6-1) = -\hat{i} - 7\hat{j} + 5\hat{k}$.
The magnitude of this vector is $|\vec{n}| = \sqrt{(-1)^2 + (-7)^2 + 5^2} = \sqrt{1 + 49 + 25} = \sqrt{75} = 5\sqrt{3}$.
The required unit vector is $\frac{\vec{n}}{|\vec{n}|} = \frac{-\hat{i} - 7\hat{j} + 5\hat{k}}{5\sqrt{3}}$.

(B) Lines $L_1$ and $L_2$ are parallel to the vectors $\vec{b}_1 = 5\hat{i} + 2\hat{j} + \hat{k}$ and $\vec{b}_2 = 4\hat{i} + 3\hat{j} + 5\hat{k}$ respectively.
The unit vector perpendicular to both $L_1$ and $L_2$ is given by $\hat{n} = \pm \frac{\vec{b}_1 \times \vec{b}_2}{|\vec{b}_1 \times \vec{b}_2|}$.
Calculating the cross product: $\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 5 & 2 & 1 \\ 4 & 3 & 5 \end{vmatrix} = \hat{i}(10-3) - \hat{j}(25-4) + \hat{k}(15-8) = 7\hat{i} - 21\hat{j} + 7\hat{k}$.
The magnitude is $|\vec{b}_1 \times \vec{b}_2| = \sqrt{7^2 + (-21)^2 + 7^2} = \sqrt{49 + 441 + 49} = \sqrt{539} = 7\sqrt{11}$.
Thus,$\hat{n} = \pm \frac{7(\hat{i} - 3\hat{j} + \hat{k})}{7\sqrt{11}} = \pm \frac{\hat{i} - 3\hat{j} + \hat{k}}{\sqrt{11}}$.
Comparing with the given options,option $B$ is correct.

(B) The lines $L_1$ and $L_2$ are parallel to the vectors $\vec{b}_1 = 3\hat{i} + \hat{j} + 2\hat{k}$ and $\vec{b}_2 = \hat{i} + 2\hat{j} + 3\hat{k}$ respectively.
The unit vector perpendicular to both $L_1$ and $L_2$ is given by $\hat{n} = \pm \frac{\vec{b}_1 \times \vec{b}_2}{|\vec{b}_1 \times \vec{b}_2|}$.
First,calculate the cross product $\vec{b}_1 \times \vec{b}_2$:
$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 2 \\ 1 & 2 & 3 \end{vmatrix} = \hat{i}(3-4) - \hat{j}(9-2) + \hat{k}(6-1) = -\hat{i} - 7\hat{j} + 5\hat{k}$.
Next,calculate the magnitude $|\vec{b}_1 \times \vec{b}_2|$:
$|\vec{b}_1 \times \vec{b}_2| = \sqrt{(-1)^2 + (-7)^2 + 5^2} = \sqrt{1 + 49 + 25} = \sqrt{75} = 5\sqrt{3}$.
Thus,the unit vector is $\hat{n} = \frac{-\hat{i} - 7\hat{j} + 5\hat{k}}{5\sqrt{3}}$.

(A) The line passes through the point with position vector $\vec{a} = 2 \hat{i} + \hat{j} - 3 \hat{k}$.
Since the line is perpendicular to the vectors $\vec{b}_1 = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b}_2 = \hat{i} + 2 \hat{j} - \hat{k}$,its direction vector $\vec{v}$ is given by the cross product $\vec{b}_1 \times \vec{b}_2$.
$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & 2 & -1 \end{vmatrix} = \hat{i}(-1 - 2) - \hat{j}(-1 - 1) + \hat{k}(2 - 1) = -3 \hat{i} + 2 \hat{j} + \hat{k}$.
The vector equation of the line is $\vec{r} = \vec{a} + \lambda \vec{v}$.
Substituting the values,we get $\vec{r} = (2 \hat{i} + \hat{j} - 3 \hat{k}) + \lambda(-3 \hat{i} + 2 \hat{j} + \hat{k})$.

(A) The normal vector to the plane $P_1$ is given by $\overline{n_1} = \overline{a} \times \overline{b}$.
The normal vector to the plane $P_2$ is given by $\overline{n_2} = \overline{c} \times \overline{d}$.
Given the condition $(\overline{a} \times \overline{b}) \times (\overline{c} \times \overline{d}) = \overline{0}$,it implies that the vector $\overline{n_1}$ is parallel to the vector $\overline{n_2}$.
Since the normal vectors of the two planes are parallel,the planes $P_1$ and $P_2$ are parallel to each other.
The angle between two parallel planes is $0$.

(B) Let the required vector be $\vec{v}$. Since $\vec{v}$ is coplanar with $\vec{a} = \hat{i}+\hat{j}+2\hat{k}$ and $\vec{b} = \hat{i}+2\hat{j}+\hat{k}$,it can be written as $\vec{v} = \vec{a} \times (\vec{a} \times \vec{b})$ or simply as a linear combination $\vec{v} = x\vec{a} + y\vec{b}$.
Alternatively,a vector coplanar to $\vec{a}$ and $\vec{b}$ and orthogonal to $\vec{c} = \hat{i}+\hat{j}+\hat{k}$ is given by $\vec{v} = \vec{c} \times (\vec{a} \times \vec{b})$.
First,calculate $\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 2 \\ 1 & 2 & 1 \end{vmatrix} = \hat{i}(1-4) - \hat{j}(1-2) + \hat{k}(2-1) = -3\hat{i} + \hat{j} + \hat{k}$.
Now,$\vec{v} = \vec{c} \times (-3\hat{i} + \hat{j} + \hat{k}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ -3 & 1 & 1 \end{vmatrix} = \hat{i}(1-1) - \hat{j}(1+3) + \hat{k}(1+3) = 0\hat{i} - 4\hat{j} + 4\hat{k}$.
The magnitude of $\vec{v}$ is $|\vec{v}| = \sqrt{0^2 + (-4)^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$.

(A) The area of a quadrilateral $ABCD$ can be calculated as the sum of the areas of two triangles,$\triangle ABC$ and $\triangle ADC$.
Area of $\triangle ABC = \frac{1}{2} |\overline{AB} \times \overline{AC}| = \frac{1}{2} |\bar{a} \times (3\bar{a} + 2\bar{b})| = \frac{1}{2} |3(\bar{a} \times \bar{a}) + 2(\bar{a} \times \bar{b})| = \frac{1}{2} |0 + 2(\bar{a} \times \bar{b})| = |\bar{a} \times \bar{b}|$.
Area of $\triangle ADC = \frac{1}{2} |\overline{AD} \times \overline{AC}| = \frac{1}{2} |\bar{b} \times (3\bar{a} + 2\bar{b})| = \frac{1}{2} |3(\bar{b} \times \bar{a}) + 2(\bar{b} \times \bar{b})| = \frac{1}{2} |-3(\bar{a} \times \bar{b}) + 0| = \frac{3}{2} |\bar{a} \times \bar{b}|$.
Total Area of quadrilateral $ABCD = |\bar{a} \times \bar{b}| + \frac{3}{2} |\bar{a} \times \bar{b}| = \frac{5}{2} |\bar{a} \times \bar{b}|$.
The area of the parallelogram with adjacent sides $\bar{a}$ and $\bar{b}$ is $|\bar{a} \times \bar{b}|$.
Given that the area of the quadrilateral is $\alpha$ times the area of the parallelogram,we have $\frac{5}{2} |\bar{a} \times \bar{b}| = \alpha |\bar{a} \times \bar{b}|$.
Therefore,$\alpha = \frac{5}{2} = 2.5$.

(C) The lines $L_1$ and $L_2$ are parallel to the vectors $\vec{b}_1 = 3 \hat{i} + 2 \hat{j} + \hat{k}$ and $\vec{b}_2 = 2 \hat{i} + \hat{j} + 3 \hat{k}$ respectively.
The unit vector perpendicular to both $L_1$ and $L_2$ is given by $\hat{n} = \frac{\vec{b}_1 \times \vec{b}_2}{|\vec{b}_1 \times \vec{b}_2|}$.
First,we calculate the cross product $\vec{b}_1 \times \vec{b}_2$:
$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 1 \\ 2 & 1 & 3 \end{vmatrix} = \hat{i}(6-1) - \hat{j}(9-2) + \hat{k}(3-4) = 5 \hat{i} - 7 \hat{j} - \hat{k}$.
Next,we find the magnitude of the cross product:
$|\vec{b}_1 \times \vec{b}_2| = \sqrt{5^2 + (-7)^2 + (-1)^2} = \sqrt{25 + 49 + 1} = \sqrt{75} = 5 \sqrt{3}$.
Therefore,the required unit vector is $\hat{n} = \frac{5 \hat{i} - 7 \hat{j} - \hat{k}}{5 \sqrt{3}}$.

(C) The projection of $\overline{a}$ on $\overline{c}$ is given by $\frac{\overline{a} \cdot \overline{c}}{|\overline{c}|}$.
Given $\overline{a} = \alpha \hat{i} + 3 \hat{j} - \hat{k}$ and $\overline{c} = \hat{i} + 2 \hat{j} - 2 \hat{k}$.
$\overline{a} \cdot \overline{c} = (\alpha)(1) + (3)(2) + (-1)(-2) = \alpha + 6 + 2 = \alpha + 8$.
$|\overline{c}| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.
So,$\frac{\alpha + 8}{3} = \frac{10}{3} \implies \alpha + 8 = 10 \implies \alpha = 2$.
Now,$\overline{b} \times \overline{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -\beta & 4 \\ 1 & 2 & -2 \end{vmatrix} = \hat{i}(2\beta - 8) - \hat{j}(-6 - 4) + \hat{k}(6 + \beta) = (2\beta - 8)\hat{i} + 10\hat{j} + (6 + \beta)\hat{k}$.
Comparing this with $-6\hat{i} + 10\hat{j} + 7\hat{k}$,we get $2\beta - 8 = -6 \implies 2\beta = 2 \implies \beta = 1$.
Finally,$2\alpha + \beta = 2(2) + 1 = 4 + 1 = 5$.

(B) Let the side vector be $\vec{a} = 3 \hat{i} + \hat{j} - \hat{k}$ and the diagonal vector be $\vec{c} = 2 \hat{i} + \hat{j} - 2 \hat{k}$.
In the parallelogram $ABCD$,let $\vec{a} = \vec{AB}$ and $\vec{c} = \vec{AC}$.
By the triangle law of vector addition in $\triangle ABC$,we have $\vec{AB} + \vec{BC} = \vec{AC}$.
Let $\vec{b} = \vec{BC}$. Then $\vec{a} + \vec{b} = \vec{c}$,which implies $\vec{b} = \vec{c} - \vec{a}$.
$\vec{b} = (2 \hat{i} + \hat{j} - 2 \hat{k}) - (3 \hat{i} + \hat{j} - \hat{k}) = -\hat{i} - \hat{k}$.
The area of the parallelogram is given by the magnitude of the cross product of the two adjacent sides,i.e.,$|\vec{a} \times \vec{b}|$.
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & -1 \\ -1 & 0 & -1 \end{vmatrix} = \hat{i}(-1 - 0) - \hat{j}(-3 - 1) + \hat{k}(0 - (-1)) = -\hat{i} + 4 \hat{j} + \hat{k}$.
Area $= |-\hat{i} + 4 \hat{j} + \hat{k}| = \sqrt{(-1)^2 + 4^2 + 1^2} = \sqrt{1 + 16 + 1} = \sqrt{18} = 3 \sqrt{2} \text{ sq. units}$.

(B) Given $\overline{a} \times \overline{b} = 2 \overline{a} \times \overline{c}$,we can write $\overline{a} \times (\overline{b} - 2 \overline{c}) = \overline{0}$.
This implies that the vector $(\overline{b} - 2 \overline{c})$ is parallel to $\overline{a}$,which is consistent with $\overline{b} - 2 \overline{c} = \lambda \overline{a}$.
Let $\alpha$ be the angle between $\overline{b}$ and $\overline{c}$.
Given $|\overline{b} \times \overline{c}| = \sqrt{15}$,we have $|\overline{b}| |\overline{c}| \sin \alpha = \sqrt{15}$.
Substituting the values,$(4)(1) \sin \alpha = \sqrt{15}$,so $\sin \alpha = \frac{\sqrt{15}}{4}$.
Then $\cos \alpha = \sqrt{1 - \sin^2 \alpha} = \sqrt{1 - \frac{15}{16}} = \sqrt{\frac{1}{16}} = \frac{1}{4}$.
Now,consider $|\overline{b} - 2 \overline{c}|^2 = |\lambda \overline{a}|^2$.
$|\overline{b}|^2 + 4|\overline{c}|^2 - 4(\overline{b} \cdot \overline{c}) = \lambda^2 |\overline{a}|^2$.
$16 + 4(1) - 4(|\overline{b}| |\overline{c}| \cos \alpha) = \lambda^2 (1)^2$.
$20 - 4(4 \times 1 \times \frac{1}{4}) = \lambda^2$.
$20 - 4 = \lambda^2$,which gives $\lambda^2 = 16$.
Thus,$\lambda = \pm 4$. Since the options provide $-4$,the correct choice is $B$.

(B) Let $\vec{a} = \hat{i}+\hat{j}+\hat{k}$ and $\vec{b} = \hat{i}+2\hat{j}+3\hat{k}$. The vector perpendicular to the plane containing $\vec{a}$ and $\vec{b}$ is given by $\vec{n} = \vec{a} \times \vec{b}$.
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & 2 & 3 \end{vmatrix} = \hat{i}(3-2) - \hat{j}(3-1) + \hat{k}(2-1) = \hat{i}-2\hat{j}+\hat{k}$.
The projection of vector $\vec{c} = 2\hat{i}+\hat{j}+\hat{k}$ on $\vec{n}$ is given by $\left| \frac{\vec{c} \cdot \vec{n}}{|\vec{n}|} \right|$.
$\vec{c} \cdot \vec{n} = (2)(1) + (1)(-2) + (1)(1) = 2 - 2 + 1 = 1$.
$|\vec{n}| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{1+4+1} = \sqrt{6}$.
Therefore,the magnitude of the projection is $\left| \frac{1}{\sqrt{6}} \right| = \frac{1}{\sqrt{6}}$.

(A) Given $\overline{a} = \hat{i} + \hat{j} + \hat{k}$ and $\overline{b} = \hat{j} - \hat{k}$. Let $\overline{r} = x \hat{i} + y \hat{j} + z \hat{k}$.
We are given $\overline{a} \times \overline{r} = \overline{b}$.
$\overline{a} \times \overline{r} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ x & y & z \end{vmatrix} = (z-y) \hat{i} - (z-x) \hat{j} + (y-x) \hat{k}$.
Comparing this with $\overline{b} = 0 \hat{i} + 1 \hat{j} - 1 \hat{k}$,we get:
$z - y = 0 \implies z = y$ $(i)$
$x - z = 1 \implies x = z + 1$ (ii)
$y - x = -1$ (iii)
Also,$\overline{a} \cdot \overline{r} = 3 \implies x + y + z = 3$ (iv).
Substituting $y = z$ and $x = z + 1$ 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,$\overline{r} = \frac{5}{3} \hat{i} + \frac{2}{3} \hat{j} + \frac{2}{3} \hat{k}$.

(B) The area of a parallelogram with adjacent sides $\bar{a}$ and $\bar{b}$ is given by $|\bar{a} \times \bar{b}|$.
Given that $|\bar{a} \times \bar{b}| = 16$.
Now,the area of the parallelogram with adjacent sides $(3 \bar{a} + 2 \bar{b})$ and $(\bar{a} + 3 \bar{b})$ is given by the magnitude of their cross product:
$|(3 \bar{a} + 2 \bar{b}) \times (\bar{a} + 3 \bar{b})|$
$= |3(\bar{a} \times \bar{a}) + 9(\bar{a} \times \bar{b}) + 2(\bar{b} \times \bar{a}) + 6(\bar{b} \times \bar{b})|$
Since $\bar{a} \times \bar{a} = 0$ and $\bar{b} \times \bar{b} = 0$,and $\bar{b} \times \bar{a} = -(\bar{a} \times \bar{b})$:
$= |0 + 9(\bar{a} \times \bar{b}) - 2(\bar{a} \times \bar{b}) + 0|$
$= |7(\bar{a} \times \bar{b})|$
$= 7 |\bar{a} \times \bar{b}|$
$= 7 \times 16 = 112$ sq. units.