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

(C) Given $a \cdot y = c$ and $a \times y = b$. Since $a$ and $b$ are perpendicular,$a \cdot b = 0$.
Taking the cross product of $a$ with the second equation:
$a \times (a \times y) = a \times b$
Using the vector triple product identity $a \times (b \times c) = (a \cdot c)b - (a \cdot b)c$:
$(a \cdot y)a - (a \cdot a)y = a \times b$
Substitute $a \cdot y = c$ and $a \cdot a = |a|^2$:
$c a - |a|^2 y = a \times b$
Rearranging for $y$:
$|a|^2 y = c a - (a \times b)$
$y = \frac{1}{|a|^2} [c a - (a \times b)]$

(A) Given $\overrightarrow{u}=\overrightarrow{a}-\overrightarrow{b}$ and $\overrightarrow{v}=\overrightarrow{a}+\overrightarrow{b}$.
$\overrightarrow{u} \times \overrightarrow{v} = (\overrightarrow{a}-\overrightarrow{b}) \times (\overrightarrow{a}+\overrightarrow{b})$
$= \overrightarrow{a} \times \overrightarrow{a} + \overrightarrow{a} \times \overrightarrow{b} - \overrightarrow{b} \times \overrightarrow{a} - \overrightarrow{b} \times \overrightarrow{b}$
Since $\overrightarrow{a} \times \overrightarrow{a} = 0$ and $\overrightarrow{b} \times \overrightarrow{b} = 0$,and $\overrightarrow{b} \times \overrightarrow{a} = -(\overrightarrow{a} \times \overrightarrow{b})$:
$= 0 + \overrightarrow{a} \times \overrightarrow{b} + \overrightarrow{a} \times \overrightarrow{b} - 0 = 2(\overrightarrow{a} \times \overrightarrow{b})$
Taking the magnitude: $|\overrightarrow{u} \times \overrightarrow{v}| = 2|\overrightarrow{a} \times \overrightarrow{b}|$
Using the identity $|\overrightarrow{a} \times \overrightarrow{b}|^2 = |\overrightarrow{a}|^2|\overrightarrow{b}|^2 - (\overrightarrow{a} \cdot \overrightarrow{b})^2$:
$|\overrightarrow{u} \times \overrightarrow{v}| = 2 \sqrt{|\overrightarrow{a}|^2|\overrightarrow{b}|^2 - (\overrightarrow{a} \cdot \overrightarrow{b})^2}$
Given $|\overrightarrow{a}| = 2$ and $|\overrightarrow{b}| = 2$,so $|\overrightarrow{a}|^2 = 4$ and $|\overrightarrow{b}|^2 = 4$:
$|\overrightarrow{u} \times \overrightarrow{v}| = 2 \sqrt{(4)(4) - (\overrightarrow{a} \cdot \overrightarrow{b})^2} = 2 \sqrt{16 - (\overrightarrow{a} \cdot \overrightarrow{b})^2}$

(A) Given $\overrightarrow{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}$.
First,we calculate the cross products:
$\overrightarrow{a} \times \hat{i} = (a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}) \times \hat{i} = a_2(\hat{j} \times \hat{i}) + a_3(\hat{k} \times \hat{i}) = -a_2 \hat{k} + a_3 \hat{j} = a_3 \hat{j} - a_2 \hat{k}$.
$\overrightarrow{a} \times \hat{j} = (a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}) \times \hat{j} = a_1(\hat{i} \times \hat{j}) + a_3(\hat{k} \times \hat{j}) = a_1 \hat{k} - a_3 \hat{i}$.
$\overrightarrow{a} \times \hat{k} = (a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}) \times \hat{k} = a_1(\hat{i} \times \hat{k}) + a_2(\hat{j} \times \hat{k}) = -a_1 \hat{j} + a_2 \hat{i} = a_2 \hat{i} - a_1 \hat{j}$.
Thus,Reason $(R)$ is true.
Now,calculate the squares of the magnitudes:
$|\overrightarrow{a} \times \hat{i}|^2 = a_3^2 + (-a_2)^2 = a_3^2 + a_2^2$.
$|\overrightarrow{a} \times \hat{j}|^2 = a_1^2 + (-a_3)^2 = a_1^2 + a_3^2$.
$|\overrightarrow{a} \times \hat{k}|^2 = a_2^2 + (-a_1)^2 = a_2^2 + a_1^2$.
Summing these gives: $|\overrightarrow{a} \times \hat{i}|^2+|\overrightarrow{a} \times \hat{j}|^2+|\overrightarrow{a} \times \hat{k}|^2 = (a_3^2 + a_2^2) + (a_1^2 + a_3^2) + (a_2^2 + a_1^2) = 2(a_1^2 + a_2^2 + a_3^2) = 2|\overrightarrow{a}|^2$.
Thus,Assertion $(A)$ is true and $(R)$ is the correct explanation for $(A)$.

(A) Given that $a$ and $b$ are unit vectors,so $|a| = 1$ and $|b| = 1$,which implies $a \cdot a = 1$ and $b \cdot b = 1$.
Expanding the cross product using the distributive property:
$(a+b) \times (a \times b) = a \times (a \times b) + b \times (a \times b)$
Using the vector triple product formula $A \times (B \times C) = (A \cdot C)B - (A \cdot B)C$:
$a \times (a \times b) = (a \cdot b)a - (a \cdot a)b = (a \cdot b)a - b$
$b \times (a \times b) = (b \cdot b)a - (b \cdot a)b = a - (a \cdot b)b$
Adding these two results:
$(a \cdot b)a - b + a - (a \cdot b)b = a - b + (a \cdot b)(a - b) = (a - b)(1 + a \cdot b)$
Since $(1 + a \cdot b)$ is a scalar,the resulting vector is parallel to $(a - b)$.

(C) The vertices of $\triangle ABC$ are $A=(2,3,5)$,$B=(-1,3,2)$,and $C=(3,5,-2)$.
First,we find the vectors $\vec{AB}$ and $\vec{AC}$:
$\vec{AB} = (-1-2)\hat{i} + (3-3)\hat{j} + (2-5)\hat{k} = -3\hat{i} - 3\hat{k}$
$\vec{AC} = (3-2)\hat{i} + (5-3)\hat{j} + (-2-5)\hat{k} = \hat{i} + 2\hat{j} - 7\hat{k}$
Now,we calculate the cross product $\vec{AB} \times \vec{AC}$:
$\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3 & 0 & -3 \\ 1 & 2 & -7 \end{vmatrix}$
$= \hat{i}(0 - (-6)) - \hat{j}(21 - (-3)) + \hat{k}(-6 - 0)$
$= 6\hat{i} - 24\hat{j} - 6\hat{k}$
The magnitude of the cross product is:
$|\vec{AB} \times \vec{AC}| = \sqrt{6^2 + (-24)^2 + (-6)^2} = \sqrt{36 + 576 + 36} = \sqrt{648} = 18\sqrt{2}$
The area of $\triangle ABC$ is given by $\frac{1}{2} |\vec{AB} \times \vec{AC}|$:
Area $= \frac{1}{2} \times 18\sqrt{2} = 9\sqrt{2} \text{ sq. units.}$

(C) Given vertices are $A=(2,3,5), B=(-1,3,2), C=(3,5,-2)$.
First,we find the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$:
$\overrightarrow{AB} = B - A = (-1-2)\hat{i} + (3-3)\hat{j} + (2-5)\hat{k} = -3\hat{i} - 3\hat{k}$
$\overrightarrow{AC} = C - A = (3-2)\hat{i} + (5-3)\hat{j} + (-2-5)\hat{k} = \hat{i} + 2\hat{j} - 7\hat{k}$
Now,calculate the cross product $\overrightarrow{AB} \times \overrightarrow{AC}$:
$\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3 & 0 & -3 \\ 1 & 2 & -7 \end{vmatrix}$
$= \hat{i}(0 - (-6)) - \hat{j}(21 - (-3)) + \hat{k}(-6 - 0)$
$= 6\hat{i} - 24\hat{j} - 6\hat{k}$
The area of $\Delta ABC$ is given by $\frac{1}{2} |\overrightarrow{AB} \times \overrightarrow{AC}|$:
$|\overrightarrow{AB} \times \overrightarrow{AC}| = \sqrt{6^2 + (-24)^2 + (-6)^2} = \sqrt{36 + 576 + 36} = \sqrt{648}$
$\sqrt{648} = \sqrt{324 \times 2} = 18\sqrt{2}$
Area $= \frac{1}{2} \times 18\sqrt{2} = 9\sqrt{2} \text{ sq. units}$.

(A) Let the direction ratios of the line $L_1$ be $\vec{a} = (1, -2, 2)$ and the direction ratios of the line $L_2$ be $\vec{b} = (-2, 3, -6)$.
To find the direction ratios of a line perpendicular to both $L_1$ and $L_2$,we calculate the cross product $\vec{n} = \vec{a} \times \vec{b}$.
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 2 \\ -2 & 3 & -6 \end{vmatrix}$
$= \hat{i}((-2)(-6) - (2)(3)) - \hat{j}((1)(-6) - (2)(-2)) + \hat{k}((1)(3) - (-2)(-2))$
$= \hat{i}(12 - 6) - \hat{j}(-6 + 4) + \hat{k}(3 - 4)$
$= \hat{i}(6) - \hat{j}(-2) + \hat{k}(-1)$
$= (6, 2, -1)$.
Thus,the direction ratios of the line perpendicular to both $L_1$ and $L_2$ are $(6, 2, -1)$.

(C) Given vectors are $a=\hat{i}+\hat{j}+\hat{k}$,$b=\hat{i}+\hat{j}+2\hat{k}$ and $c=2\hat{i}+3\hat{j}+4\hat{k}$.
First,we find the vector perpendicular to both $a$ and $b$ using the cross product:
$a \times b = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & 1 & 2 \end{vmatrix} = \hat{i}(2-1) - \hat{j}(2-1) + \hat{k}(1-1) = \hat{i} - \hat{j}$.
The unit vector perpendicular to both $a$ and $b$ is given by $n = \pm \frac{\hat{i} - \hat{j}}{|\hat{i} - \hat{j}|} = \pm \frac{\hat{i} - \hat{j}}{\sqrt{1^2 + (-1)^2}} = \pm \frac{\hat{i} - \hat{j}}{\sqrt{2}}$.
The magnitude of the projection of vector $n$ on vector $c$ is given by $|n \cdot \hat{c}|$,where $\hat{c} = \frac{c}{|c|}$.
$|c| = \sqrt{2^2 + 3^2 + 4^2} = \sqrt{4 + 9 + 16} = \sqrt{29}$.
Projection magnitude $= \left| \left( \pm \frac{\hat{i} - \hat{j}}{\sqrt{2}} \right) \cdot \left( \frac{2\hat{i} + 3\hat{j} + 4\hat{k}}{\sqrt{29}} \right) \right| = \left| \frac{\pm(2 - 3)}{\sqrt{2}\sqrt{29}} \right| = \left| \frac{-1}{\sqrt{58}} \right| = \frac{1}{\sqrt{58}}$.
Thus,the correct option is $C$.

(B) Let the direction ratios of the line perpendicular to both $L_1$ and $L_2$ be $(a, b, c)$. Since the line is perpendicular to both,its direction ratios are given by the cross product of the direction ratios of $L_1$ and $L_2$:
$(a, b, c) = (2, -1, 1) \times (3, -3, 4) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 3 & -3 & 4 \end{vmatrix} = \hat{i}(-4 + 3) - \hat{j}(8 - 3) + \hat{k}(-6 + 3) = -1\hat{i} - 5\hat{j} - 3\hat{k}$.
Thus,the direction ratios are $(-1, -5, -3)$ or $(1, 5, 3)$.
The magnitude is $\sqrt{1^2 + 5^2 + 3^2} = \sqrt{1 + 25 + 9} = \sqrt{35}$.
The direction cosines are $\pm \frac{1}{\sqrt{35}}, \pm \frac{5}{\sqrt{35}}, \pm \frac{3}{\sqrt{35}}$.

(D) The normal $\vec{n}_1$ to the plane $P_1$ determined by $\hat{j}-\hat{k}$ and $3\hat{j}-2\hat{k}$ is given by the cross product:
$\vec{n}_1 = (\hat{j}-\hat{k}) \times (3\hat{j}-2\hat{k}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & -1 \\ 0 & 3 & -2 \end{vmatrix} = \hat{i}(-2+3) = \hat{i}$.
The normal $\vec{n}_2$ to the plane $P_2$ determined by $2\hat{i}+3\hat{j}$ and $\hat{i}-3\hat{j}$ is given by:
$\vec{n}_2 = (2\hat{i}+3\hat{j}) \times (\hat{i}-3\hat{j}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 0 \\ 1 & -3 & 0 \end{vmatrix} = \hat{k}(-6-3) = -9\hat{k}$.
Since $\vec{a}$ is parallel to the line of intersection of planes $P_1$ and $P_2$,it is parallel to $\vec{n}_1 \times \vec{n}_2$:
$\vec{a} = k(\vec{n}_1 \times \vec{n}_2) = k(\hat{i} \times -9\hat{k}) = k(9\hat{j}) = 9k\hat{j}$.
We can take $\vec{a} = \pm\hat{j}$.
The angle $\theta$ between $\vec{a} = \hat{j}$ and $\vec{b} = \hat{i}+\hat{j}+\hat{k}$ is given by:
$\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{1}{1 \cdot \sqrt{1^2+1^2+1^2}} = \frac{1}{\sqrt{3}}$.
Considering the direction $\vec{a} = -\hat{j}$,we get $\cos \theta = -\frac{1}{\sqrt{3}}$.
Thus,$\theta = \cos^{-1}\left(\pm\frac{1}{\sqrt{3}}\right)$.

(B) Let the required vector be $\vec{v} = a(2 \hat{i}+\hat{j}+\hat{k}) + b(\hat{i}-\hat{j}+\hat{k})$.
This simplifies to $\vec{v} = (2a+b)\hat{i} + (a-b)\hat{j} + (a+b)\hat{k}$.
Since $\vec{v}$ is orthogonal to $3 \hat{i}+2 \hat{j}+6 \hat{k}$,their dot product is zero:
$3(2a+b) + 2(a-b) + 6(a+b) = 0$.
$6a + 3b + 2a - 2b + 6a + 6b = 0$.
$14a + 7b = 0 \implies b = -2a$.
Substituting $b = -2a$ into the expression for $\vec{v}$:
$\vec{v} = (2a - 2a)\hat{i} + (a - (-2a))\hat{j} + (a + (-2a))\hat{k} = 0\hat{i} + 3a\hat{j} - a\hat{k} = a(3\hat{j} - \hat{k})$.
The unit vector is $\pm \frac{\vec{v}}{|\vec{v}|} = \pm \frac{a(3\hat{j} - \hat{k})}{|a|\sqrt{3^2 + (-1)^2}} = \pm \frac{3\hat{j} - \hat{k}}{\sqrt{10}}$.

(C) Let the diagonals of the parallelogram be $\overrightarrow{d}_1 = 2 \hat{i} + 3 \hat{j} - 4 \hat{k}$ and $\overrightarrow{d}_2 = -\hat{i} + 2 \hat{j} + \hat{k}$.
The area of a parallelogram with diagonals $\overrightarrow{d}_1$ and $\overrightarrow{d}_2$ is given by the formula: $\text{Area} = \frac{1}{2} |\overrightarrow{d}_1 \times \overrightarrow{d}_2|$.
First,calculate the cross product $\overrightarrow{d}_1 \times \overrightarrow{d}_2$:
$\overrightarrow{d}_1 \times \overrightarrow{d}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -4 \\ -1 & 2 & 1 \end{vmatrix}$
$= \hat{i}(3(1) - (-4)(2)) - \hat{j}(2(1) - (-4)(-1)) + \hat{k}(2(2) - 3(-1))$
$= \hat{i}(3 + 8) - \hat{j}(2 - 4) + \hat{k}(4 + 3)$
$= 11 \hat{i} + 2 \hat{j} + 7 \hat{k}$.
Now,calculate the magnitude of the cross product:
$|\overrightarrow{d}_1 \times \overrightarrow{d}_2| = \sqrt{11^2 + 2^2 + 7^2} = \sqrt{121 + 4 + 49} = \sqrt{174}$.
Finally,the area is $\frac{1}{2} \sqrt{174}$.

(A) Let the position vectors of vertices $A, B,$ and $C$ be $\vec{a} = 3\hat{i}+4\hat{j}-\hat{k}$,$\vec{b} = \hat{i}+3\hat{j}+\hat{k}$,and $\vec{c} = 5\hat{i}+5\hat{j}+5\hat{k}$.
The vectors representing the sides are:
$\vec{AB} = \vec{b} - \vec{a} = (1-3)\hat{i} + (3-4)\hat{j} + (1-(-1))\hat{k} = -2\hat{i} - \hat{j} + 2\hat{k}$
$\vec{AC} = \vec{c} - \vec{a} = (5-3)\hat{i} + (5-4)\hat{j} + (5-(-1))\hat{k} = 2\hat{i} + \hat{j} + 6\hat{k}$
$\vec{BC} = \vec{c} - \vec{b} = (5-1)\hat{i} + (5-3)\hat{j} + (5-1)\hat{k} = 4\hat{i} + 2\hat{j} + 4\hat{k}$
The area of $\triangle ABC$ is given by $\frac{1}{2} |\vec{AB} \times \vec{AC}|$.
$\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & -1 & 2 \\ 2 & 1 & 6 \end{vmatrix} = \hat{i}(-6-2) - \hat{j}(-12-4) + \hat{k}(-2+2) = -8\hat{i} + 16\hat{j} + 0\hat{k}$
Magnitude $|\vec{AB} \times \vec{AC}| = \sqrt{(-8)^2 + 16^2 + 0^2} = \sqrt{64 + 256} = \sqrt{320} = 8\sqrt{5}$.
Area $= \frac{1}{2} \times 8\sqrt{5} = 4\sqrt{5}$.
Also,Area $= \frac{1}{2} \times |\vec{BC}| \times p$,where $p$ is the altitude from $A$ to $BC$.
$|\vec{BC}| = \sqrt{4^2 + 2^2 + 4^2} = \sqrt{16+4+16} = \sqrt{36} = 6$.
So,$4\sqrt{5} = \frac{1}{2} \times 6 \times p \implies 4\sqrt{5} = 3p \implies p = \frac{4\sqrt{5}}{3}$.

(A) Given that $a, b, c, d$ are coplanar vectors.
Since $a$ and $b$ are coplanar,the vector $a \times b$ is perpendicular to the plane containing $a$ and $b$.
Similarly,since $c$ and $d$ are coplanar,the vector $c \times d$ is perpendicular to the plane containing $c$ and $d$.
Because $a, b, c, d$ all lie in the same plane,the vectors $a \times b$ and $c \times d$ are both perpendicular to the same plane.
Therefore,$a \times b$ and $c \times d$ are parallel to each other.
Since the cross product of two parallel vectors is the zero vector,we have $(a \times b) \times (c \times d) = 0$.

(D) The vector $\vec{r}$ is perpendicular to the plane determined by $\vec{a} = 2 \hat{i} - \hat{j}$ and $\vec{b} = \hat{j} + 2 \hat{k}$. Thus,$\vec{r}$ is parallel to $\vec{a} \times \vec{b}$.
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 0 \\ 0 & 1 & 2 \end{vmatrix} = \hat{i}(-2 - 0) - \hat{j}(4 - 0) + \hat{k}(2 - 0) = -2 \hat{i} - 4 \hat{j} + 2 \hat{k}$.
Let $\vec{r} = \lambda(-2 \hat{i} - 4 \hat{j} + 2 \hat{k}) = 2\lambda(-\hat{i} - 2 \hat{j} + \hat{k})$.
The projection of $\vec{r}$ on $\vec{v} = 2 \hat{i} + \hat{j} + 2 \hat{k}$ is given by $\frac{|\vec{r} \cdot \vec{v}|}{|\vec{v}|} = 1$.
$|\vec{v}| = \sqrt{2^2 + 1^2 + 2^2} = \sqrt{4 + 1 + 4} = 3$.
$\vec{r} \cdot \vec{v} = \lambda(-2 \hat{i} - 4 \hat{j} + 2 \hat{k}) \cdot (2 \hat{i} + \hat{j} + 2 \hat{k}) = \lambda(-4 - 4 + 4) = -4\lambda$.
So,$\frac{|-4\lambda|}{3} = 1 \Rightarrow |\lambda| = \frac{3}{4}$.
$|\vec{r}| = |\lambda| \sqrt{(-2)^2 + (-4)^2 + 2^2} = \frac{3}{4} \sqrt{4 + 16 + 4} = \frac{3}{4} \sqrt{24} = \frac{3}{4} \times 2 \sqrt{6} = \frac{3 \sqrt{6}}{2}$.

(B) Given $\vec{a}=\hat{i}+2 \hat{j}+\hat{k}$ and $\vec{b}=3 \hat{i}-3 \hat{j}+3 \hat{k}$.
Let $\vec{c}=x \hat{i}+y \hat{j}+z \hat{k}$.
From $\vec{a} \times \vec{c}=\vec{b}$,we have:
$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 1 \\ x & y & z \end{vmatrix} = 3 \hat{i}-3 \hat{j}+3 \hat{k}$
$(2z-y) \hat{i} - (z-x) \hat{j} + (y-2x) \hat{k} = 3 \hat{i}-3 \hat{j}+3 \hat{k}$
Comparing components:
$2z-y=3$ $(i)$
$x-z=-3 \Rightarrow z-x=3$ (ii)
$y-2x=3$ (iii)
Given $\vec{a} \cdot \vec{c}=3$,so $x+2y+z=3$ (iv).
From (ii),$z=x+3$. Substituting into (iv): $x+2y+x+3=3 \Rightarrow 2x+2y=0 \Rightarrow y=-x$.
Substituting $y=-x$ into (iii): $-x-2x=3 \Rightarrow -3x=3 \Rightarrow x=-1$.
Then $y=1$ and $z=2$. Thus,$\vec{c}=-\hat{i}+\hat{j}+2 \hat{k}$.
Now,$\vec{c} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & 2 \\ 3 & -3 & 3 \end{vmatrix} = (3+6) \hat{i} - (-3-6) \hat{j} + (3-3) \hat{k} = 9 \hat{i}+9 \hat{j}$.
We need to calculate $\vec{a} \cdot (\vec{c} \times \vec{b} - \vec{b} - \vec{c}) = \vec{a} \cdot (\vec{c} \times \vec{b}) - \vec{a} \cdot \vec{b} - \vec{a} \cdot \vec{c}$.
$\vec{a} \cdot (\vec{c} \times \vec{b}) = (\hat{i}+2 \hat{j}+\hat{k}) \cdot (9 \hat{i}+9 \hat{j}) = 9+18=27$.
$\vec{a} \cdot \vec{b} = (1)(3) + (2)(-3) + (1)(3) = 3-6+3=0$.
$\vec{a} \cdot \vec{c} = 3$.
Therefore,$27 - 0 - 3 = 24$.

(A) Given $|\vec{a}|=|\vec{b}|=|\vec{c}|=\sqrt{2}$ and the angle between any two vectors is $\frac{\pi}{3}$.
$\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos(\frac{\pi}{3}) = \sqrt{2} \times \sqrt{2} \times \frac{1}{2} = 1$.
Similarly,$\vec{b} \cdot \vec{c} = 1$ and $\vec{c} \cdot \vec{a} = 1$.
Using the vector triple product formula $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$:
$\vec{x} = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c} = 1 \cdot \vec{b} - 1 \cdot \vec{c} = \vec{b} - \vec{c}$.
Now,$|\vec{x}|^2 = |\vec{b} - \vec{c}|^2 = |\vec{b}|^2 + |\vec{c}|^2 - 2(\vec{b} \cdot \vec{c}) = 2 + 2 - 2(1) = 2$.
Similarly,$\vec{y} = \vec{b} \times (\vec{c} \times \vec{a}) = (\vec{b} \cdot \vec{a})\vec{c} - (\vec{b} \cdot \vec{c})\vec{a} = 1 \cdot \vec{c} - 1 \cdot \vec{a} = \vec{c} - \vec{a}$.
Now,$|\vec{y}|^2 = |\vec{c} - \vec{a}|^2 = |\vec{c}|^2 + |\vec{a}|^2 - 2(\vec{c} \cdot \vec{a}) = 2 + 2 - 2(1) = 2$.
Since $|\vec{x}|^2 = 2$ and $|\vec{y}|^2 = 2$,we have $|\vec{x}| = |\vec{y}| = \sqrt{2}$.

(C) Since $\vec{d}$ is normal to the plane of $\vec{a}$ and $\vec{b}$,it must be parallel to $\vec{a} \times \vec{b}$.
Thus,$\vec{d} = \lambda(\vec{a} \times \vec{b})$.
First,calculate the cross product $\vec{a} \times \vec{b}$:
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -2 \\ 1 & -2 & 1 \end{vmatrix} = \hat{i}(1 - 4) - \hat{j}(1 - (-2)) + \hat{k}(-2 - 1) = -3\hat{i} - 3\hat{j} - 3\hat{k}$.
We can write $\vec{d} = \mu(\hat{i} + \hat{j} + \hat{k})$ where $\mu = -3\lambda$.
Given $\vec{d} \cdot \vec{c} = 2$,we substitute $\vec{c} = 2\hat{i} + \hat{j} - \hat{k}$:
$\mu(\hat{i} + \hat{j} + \hat{k}) \cdot (2\hat{i} + \hat{j} - \hat{k}) = 2$
$\mu(2 + 1 - 1) = 2 \Rightarrow 2\mu = 2 \Rightarrow \mu = 1$.
Therefore,$\vec{d} = \hat{i} + \hat{j} + \hat{k}$.
The magnitude is $|\vec{d}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$.

(A) Let $\vec{a} = x \hat{i} + y \hat{j} + z \hat{k}$.
Given $\vec{a} \times \hat{i} = \hat{j} + \hat{k}$.
Calculating the cross product: $\vec{a} \times \hat{i} = (x \hat{i} + y \hat{j} + z \hat{k}) \times \hat{i} = -y \hat{k} + z \hat{j} = z \hat{j} - y \hat{k}$.
Comparing with $\hat{j} + \hat{k}$,we get $z = 1$ and $y = -1$.
Given $\vec{a} \cdot \hat{i} = 1$,so $(x \hat{i} + y \hat{j} + z \hat{k}) \cdot \hat{i} = x = 1$.
Thus,$\vec{a} = \hat{i} - \hat{j} + \hat{k}$.
The equation of the line passing through point $\vec{p} = \hat{i} + \hat{j} + \hat{k}$ and parallel to $\vec{a}$ is $\vec{r} = \vec{p} + t \vec{a}$.
$\vec{r} = (\hat{i} + \hat{j} + \hat{k}) + t(\hat{i} - \hat{j} + \hat{k}) = (1+t) \hat{i} + (1-t) \hat{j} + (1+t) \hat{k}$.

(A) Given $\vec{p} = \vec{q} + \vec{r}$,we have:
$a \hat{i} + b \hat{j} + c \hat{k} = (d \hat{i} + 3 \hat{j} + 4 \hat{k}) + (3 \hat{i} + \hat{j} - 2 \hat{k})$
$a \hat{i} + b \hat{j} + c \hat{k} = (d + 3) \hat{i} + 4 \hat{j} + 2 \hat{k}$
Comparing coefficients,we get $a = d + 3 \Rightarrow d = a - 3$,$b = 4$,and $c = 2$.
The area of $\triangle ABC$ formed by vectors $\vec{q}$ and $\vec{r}$ is given by $\frac{1}{2} |\vec{q} \times \vec{r}| = 5 \sqrt{6}$.
$\vec{q} \times \vec{r} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ d & 3 & 4 \\ 3 & 1 & -2 \end{vmatrix} = \hat{i}(-6 - 4) - \hat{j}(-2d - 12) + \hat{k}(d - 9) = -10 \hat{i} + (2d + 12) \hat{j} + (d - 9) \hat{k}$.
Area $= \frac{1}{2} \sqrt{(-10)^2 + (2d + 12)^2 + (d - 9)^2} = 5 \sqrt{6}$.
$\sqrt{100 + 4d^2 + 48d + 144 + d^2 - 18d + 81} = 10 \sqrt{6}$.
$5d^2 + 30d + 325 = 600 \Rightarrow 5d^2 + 30d - 275 = 0 \Rightarrow d^2 + 6d - 55 = 0$.
$(d + 11)(d - 5) = 0$,so $d = 5$ or $d = -11$.
If $d = 5$,$a = 5 + 3 = 8$. Then $|a| + |b| + |c| = 8 + 4 + 2 = 14$.
If $d = -11$,$a = -11 + 3 = -8$. Then $|a| + |b| + |c| = |-8| + 4 + 2 = 14$.

(B) vector perpendicular to both $\vec{b}$ and $\vec{c}$ is given by $\vec{r} = \lambda(\vec{b} \times \vec{c})$.
$\vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 5 \\ 1 & -4 & -2 \end{vmatrix} = \hat{i}(2 + 20) - \hat{j}(-6 - 5) + \hat{k}(-12 + 1) = 22\hat{i} + 11\hat{j} - 11\hat{k} = 11(2\hat{i} + \hat{j} - \hat{k})$.
Thus,$\vec{r} = \lambda(11)(2\hat{i} + \hat{j} - \hat{k})$.
Given $\vec{r} \cdot \vec{a} = 11$,we have $11\lambda(2\hat{i} + \hat{j} - \hat{k}) \cdot (\hat{i} + 2\hat{j} + 3\hat{k}) = 11$.
$11\lambda(2(1) + 1(2) - 1(3)) = 11 \implies 11\lambda(2 + 2 - 3) = 11 \implies 11\lambda(1) = 11 \implies \lambda = 1$.
So,$\vec{r} = 11(2\hat{i} + \hat{j} - \hat{k})$.
$A$ vector $\vec{p}$ is perpendicular to $\vec{r}$ if $\vec{r} \cdot \vec{p} = 0$,which implies $(2\hat{i} + \hat{j} - \hat{k}) \cdot \vec{p} = 0$.
Checking option $B$: $(2\hat{i} + \hat{j} - \hat{k}) \cdot (\hat{i} - \hat{j} + \hat{k}) = 2(1) + 1(-1) - 1(1) = 2 - 1 - 1 = 0$.
Thus,the vector $\hat{i} - \hat{j} + \hat{k}$ is perpendicular to $\vec{r}$.

(D) First,calculate the cross product $\vec{a} \times \vec{b}$:
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & 1 \\ 1 & 2 & -3 \end{vmatrix} = \hat{i}(9-2) - \hat{j}(-6-1) + \hat{k}(4+3) = 7\hat{i} + 7\hat{j} + 7\hat{k}$
Next,calculate $(\vec{a} \times \vec{b}) \times \vec{c}$:
$(\vec{a} \times \vec{b}) \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 7 & 7 & 7 \\ 1 & -1 & 0 \end{vmatrix} = \hat{i}(0 - (-7)) - \hat{j}(0 - 7) + \hat{k}(-7 - 7) = 7\hat{i} + 7\hat{j} - 14\hat{k}$
Since $(\vec{a} \times \vec{b}) \times \vec{c}$ is perpendicular to $\vec{d}$,their dot product must be zero:
$(7\hat{i} + 7\hat{j} - 14\hat{k}) \cdot (\hat{i} + \hat{j} + x\hat{k}) = 0$
$7(1) + 7(1) - 14(x) = 0$
$14 - 14x = 0$
$14x = 14$
$x = 1$

(C) Given,$\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{c}=\hat{j}-\hat{k}$.
Let $\vec{b}=x\hat{i}+y\hat{j}+z\hat{k}$.
From $\vec{a} \cdot \vec{b}=3$,we have $(\hat{i}+\hat{j}+\hat{k}) \cdot (x\hat{i}+y\hat{j}+z\hat{k})=3$,which implies $x+y+z=3$ ... $(1)$.
From $\vec{a} \times \vec{b}=\vec{c}$,we have:
$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ x & y & z \end{vmatrix} = \hat{j}-\hat{k}$
$\hat{i}(z-y) - \hat{j}(z-x) + \hat{k}(y-x) = 0\hat{i} + 1\hat{j} - 1\hat{k}$.
Comparing coefficients:
$z-y=0 \Rightarrow z=y$
$-(z-x)=1 \Rightarrow x-z=1 \Rightarrow x=z+1$
$y-x=-1 \Rightarrow x=y+1$.
Substituting $y=z$ and $x=z+1$ into $(1)$:
$(z+1) + z + z = 3 \Rightarrow 3z+1=3 \Rightarrow 3z=2 \Rightarrow z=\frac{2}{3}$.
Thus,$y=\frac{2}{3}$ and $x=\frac{2}{3}+1=\frac{5}{3}$.
Therefore,$\vec{b}=\frac{5}{3}\hat{i}+\frac{2}{3}\hat{j}+\frac{2}{3}\hat{k}$.

(D) Let $\vec{a}$ be the required unit vector. Since $\vec{a}$ is coplanar with $\vec{c} = \hat{i}+\hat{j}+\hat{k}$ and $\vec{d} = 2 \hat{i}-\hat{j}-\hat{k}$,it must be perpendicular to the vector $\vec{n} = \vec{c} \times \vec{d}$.
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & -1 & -1 \end{vmatrix} = \hat{i}(-1+1) - \hat{j}(-1-2) + \hat{k}(-1-2) = 0\hat{i} + 3\hat{j} - 3\hat{k}$.
Since $\vec{a}$ is also perpendicular to $\vec{b} = \hat{i}-2 \hat{j}+3 \hat{k}$,$\vec{a}$ must be parallel to $\vec{b} \times \vec{n}$.
$\vec{b} \times \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 0 & 3 & -3 \end{vmatrix} = \hat{i}(6-9) - \hat{j}(-3-0) + \hat{k}(3-0) = -3\hat{i} + 3\hat{j} + 3\hat{k}$.
The unit vector is $\pm \frac{\vec{b} \times \vec{n}}{|\vec{b} \times \vec{n}|} = \pm \frac{-3\hat{i} + 3\hat{j} + 3\hat{k}}{\sqrt{(-3)^2 + 3^2 + 3^2}} = \pm \frac{-3\hat{i} + 3\hat{j} + 3\hat{k}}{\sqrt{27}} = \pm \frac{-3\hat{i} + 3\hat{j} + 3\hat{k}}{3\sqrt{3}} = \pm \frac{1}{\sqrt{3}}(-\hat{i} + \hat{j} + \hat{k})$.
Note: The vector $\pm \frac{1}{\sqrt{3}}(\hat{i}-\hat{j}-\hat{k})$ is equivalent to $\pm \frac{1}{\sqrt{3}}(-\hat{i}+\hat{j}+\hat{k})$.

(D) The area of a triangle with sides $\vec{a}$ and $\vec{b}$ is given by the formula: $\text{Area} = \frac{1}{2} |\vec{a} \times \vec{b}|$.
Since $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta$,where $\theta$ is the angle between the vectors:
First,calculate the magnitude of $\vec{a}$: $|\vec{a}| = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.
Given $|\vec{b}| = 6$ and $\theta = \frac{\pi}{6}$.
Substitute these values into the area formula: $\text{Area} = \frac{1}{2} \times |\vec{a}| \times |\vec{b}| \times \sin \left(\frac{\pi}{6}\right)$.
$\text{Area} = \frac{1}{2} \times 3 \times 6 \times \sin \left(\frac{\pi}{6}\right)$.
Since $\sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$,we have:
$\text{Area} = \frac{1}{2} \times 3 \times 6 \times \frac{1}{2} = \frac{18}{4} = \frac{9}{2}$ sq. units.

(A) Let $b=x \hat{i}+y \hat{j}+z \hat{k}$.
The cross product $a \times b$ is given by the determinant:
$a \times b = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ x & y & z \end{vmatrix} = \hat{i}(z-y) - \hat{j}(z-x) + \hat{k}(y-x) = (z-y) \hat{i} + (x-z) \hat{j} + (y-x) \hat{k}$.
Given $a \times b = c = 0 \hat{i} + 1 \hat{j} - 1 \hat{k}$,we equate the components:
$z-y = 0 \Rightarrow y=z$
$x-z = 1 \Rightarrow z=x-1$
$y-x = -1 \Rightarrow y=x-1$.
Also,given $a \cdot b = 3$,we have $x+y+z = 3$.
Substituting $y$ and $z$ in terms of $x$:
$x + (x-1) + (x-1) = 3$
$3x - 2 = 3 \Rightarrow 3x = 5 \Rightarrow x = \frac{5}{3}$.
Then $y = z = \frac{5}{3} - 1 = \frac{2}{3}$.
Thus,$b = \frac{5}{3} \hat{i} + \frac{2}{3} \hat{j} + \frac{2}{3} \hat{k} = \frac{1}{3}(5 \hat{i} + 2 \hat{j} + 2 \hat{k})$.

(B) Given vectors are $a=2 \hat{i}-3 \hat{j}+4 \hat{k}$,$b=7 \hat{i}+2 \hat{j}-3 \hat{k}$,and $c=\hat{i}+\hat{j}+\hat{k}$.
Since $x$ is perpendicular to both $a$ and $b$,$x$ must be parallel to the cross product $a \times b$.
Thus,$x = \lambda(a \times b)$.
Calculating the cross product $a \times b$:
$a \times b = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & 4 \\ 7 & 2 & -3 \end{vmatrix} = \hat{i}(9-8) - \hat{j}(-6-28) + \hat{k}(4+21) = \hat{i} + 34 \hat{j} + 25 \hat{k}$.
So,$x = \lambda(\hat{i} + 34 \hat{j} + 25 \hat{k})$.
Given $x \cdot c = 60$,we substitute $x$ and $c$:
$(\lambda \hat{i} + 34 \lambda \hat{j} + 25 \lambda \hat{k}) \cdot (\hat{i} + \hat{j} + \hat{k}) = 60$.
$\lambda + 34 \lambda + 25 \lambda = 60$.
$60 \lambda = 60$,which gives $\lambda = 1$.
Therefore,$x = \hat{i} + 34 \hat{j} + 25 \hat{k}$.

(A) Let $\vec{a}$ make an angle $\alpha$ with the $Z$-axis and $\vec{b}$ make an angle $\beta$ with the $X$-axis. Using the direction cosines property $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$:
For vector $\vec{a}$: $\cos^2 60^{\circ} + \cos^2 60^{\circ} + \cos^2 \alpha = 1 \Rightarrow \frac{1}{4} + \frac{1}{4} + \cos^2 \alpha = 1 \Rightarrow \cos^2 \alpha = \frac{1}{2} \Rightarrow \cos \alpha = \pm \frac{1}{\sqrt{2}}$.
Thus,$\vec{a} = 2(\cos 60^{\circ} \hat{i} + \cos 60^{\circ} \hat{j} + \cos \alpha \hat{k}) = 2(\frac{1}{2} \hat{i} + \frac{1}{2} \hat{j} \pm \frac{1}{\sqrt{2}} \hat{k}) = \hat{i} + \hat{j} \pm \sqrt{2} \hat{k}$.
For vector $\vec{b}$: $\cos^2 \beta + \cos^2 45^{\circ} + \cos^2 45^{\circ} = 1 \Rightarrow \cos^2 \beta + \frac{1}{2} + \frac{1}{2} = 1 \Rightarrow \cos \beta = 0$.
Thus,$\vec{b} = \sqrt{2}(0 \hat{i} + \cos 45^{\circ} \hat{j} + \cos 45^{\circ} \hat{k}) = \sqrt{2}(0 \hat{i} + \frac{1}{\sqrt{2}} \hat{j} + \frac{1}{\sqrt{2}} \hat{k}) = \hat{j} + \hat{k}$.
Now,$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & \pm \sqrt{2} \\ 0 & 1 & 1 \end{vmatrix} = \hat{i}(1 - (\pm \sqrt{2})) - \hat{j}(1 - 0) + \hat{k}(1 - 0) = (1 \mp \sqrt{2}) \hat{i} - \hat{j} + \hat{k}$.
Taking the positive sign,we get $(1 - \sqrt{2}) \hat{i} - \hat{j} + \hat{k}$.

(A) Let $C$ be a point on the line such that $\overrightarrow{BC}$ is parallel to $\overrightarrow{OA} \times \overrightarrow{OB}$ and $|\overrightarrow{BC}| = \sqrt{189}$.
First,calculate the cross product $\vec{v} = \overrightarrow{OA} \times \overrightarrow{OB}$:
$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 4 & 0 & 1 \end{vmatrix} = \hat{i}(2-0) - \hat{j}(1-12) + \hat{k}(0-8) = 2\hat{i} + 11\hat{j} - 8\hat{k}$.
Since $\overrightarrow{BC}$ is parallel to $\vec{v}$,we have $\overrightarrow{BC} = \lambda(2\hat{i} + 11\hat{j} - 8\hat{k})$.
The magnitude is $|\overrightarrow{BC}| = |\lambda| \sqrt{2^2 + 11^2 + (-8)^2} = |\lambda| \sqrt{4 + 121 + 64} = |\lambda| \sqrt{189}$.
Given $|\overrightarrow{BC}| = \sqrt{189}$,we get $|\lambda| = 1$,so $\lambda = \pm 1$.
The position vector of $C$ is $\vec{OC} = \vec{OB} + \overrightarrow{BC} = (4\hat{i} + \hat{k}) + \lambda(2\hat{i} + 11\hat{j} - 8\hat{k})$.
For $\lambda = 1$,$\vec{OC} = (4+2)\hat{i} + (0+11)\hat{j} + (1-8)\hat{k} = 6\hat{i} + 11\hat{j} - 7\hat{k}$.

(B) First,calculate the cross product $a \times b$:
$a \times b = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -3 \\ 1 & -2 & 3 \end{vmatrix} = \hat{i}(3-6) - \hat{j}(6+3) + \hat{k}(-4-1) = -3 \hat{i} - 9 \hat{j} - 5 \hat{k}$.
Next,calculate the cross product $c \times d$:
$c \times d = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & -4 \\ 1 & 1 & 2 \end{vmatrix} = \hat{i}(2+4) - \hat{j}(-2+4) + \hat{k}(-1-1) = 6 \hat{i} - 2 \hat{j} - 2 \hat{k}$.
Finally,calculate the cross product of the two resulting vectors:
$(a \times b) \times (c \times d) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3 & -9 & -5 \\ 6 & -2 & -2 \end{vmatrix} = \hat{i}(18-10) - \hat{j}(6+30) + \hat{k}(6+54) = 8 \hat{i} - 36 \hat{j} + 60 \hat{k}$.

(D) Given equations are $r \times a = b \times a$ and $r \times b = a \times b$.
From the first equation,$r \times a - b \times a = 0$,which implies $(r - b) \times a = 0$.
From the second equation,$r \times b - a \times b = 0$,which implies $(r - a) \times b = 0$.
Adding the two equations: $r \times a + r \times b = b \times a + a \times b$.
Since $b \times a = -(a \times b)$,we have $r \times (a + b) = 0$.
This implies that $r$ is parallel to $(a + b)$.
Thus,$r = k(a + b)$ for some scalar $k$.
Substituting $r = a + b$ into the original equations:
$(a + b) \times a = a \times a + b \times a = 0 + b \times a = b \times a$ (Satisfied).
$(a + b) \times b = a \times b + b \times b = a \times b + 0 = a \times b$ (Satisfied).
Therefore,$r = a + b = (\hat{i} + \hat{j}) + (3 \hat{i} - 2 \hat{j}) = 4 \hat{i} - \hat{j}$.

(C) Given that $a, b, c$ are unit vectors,so $|a|=|b|=|c|=1$.
Given $a+b+c=0$.
The angle between $a$ and $b$ is $\frac{\pi}{3}$.
Taking the cross product of $a+b+c=0$ with $a$:
$a \times (a+b+c) = a \times 0$
$a \times a + a \times b + a \times c = 0$
Since $a \times a = 0$,we have $a \times b = c \times a$.
Taking the magnitude,$|a \times b| = |c \times a|$.
Similarly,by taking the cross product with $b$,we get $|a \times b| = |b \times c|$.
Thus,$|a \times b| = |b \times c| = |c \times a|$.
Therefore,$|a \times b| + |b \times c| + |c \times a| = 3|a \times b|$.
Using the formula $|a \times b| = |a||b| \sin(\theta)$,where $\theta = \frac{\pi}{3}$:
$|a \times b| = 1 \times 1 \times \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$.
So,$3|a \times b| = 3 \times \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2}$.

(D) Given: $|a|=1, |b|=2$ and the angle $\theta = 120^{\circ}$.
We need to evaluate the expression ${(a+3b) \times (3a-b)}^2$.
First,expand the cross product:
$(a+3b) \times (3a-b) = a \times (3a) - a \times b + (3b) \times (3a) - (3b) \times b$
Since $a \times a = 0$ and $b \times b = 0$,we have:
$= 0 - (a \times b) + 9(b \times a) - 0$
Since $b \times a = -(a \times b)$,the expression becomes:
$= -(a \times b) - 9(a \times b) = -10(a \times b)$
Now,square the magnitude:
${(-10)(a \times b)}^2 = 100 |a \times b|^2$
Using the formula $|a \times b| = |a||b| \sin \theta$:
$|a \times b| = 1 \times 2 \times \sin 120^{\circ} = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3}$
Therefore,$100 |a \times b|^2 = 100 \times (\sqrt{3})^2 = 100 \times 3 = 300$.

(A) Given $\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}$,$\overrightarrow{b}=\hat{i}+\hat{j}$,$\overrightarrow{c}=\hat{i}$.
First,calculate the cross product $\overrightarrow{a} \times \overrightarrow{b}$:
$\overrightarrow{a} \times \overrightarrow{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & 1 & 0 \end{vmatrix} = \hat{i}(0-1) - \hat{j}(0-1) + \hat{k}(1-1) = -\hat{i} + \hat{j}$.
Now,calculate $(\overrightarrow{a} \times \overrightarrow{b}) \times \overrightarrow{c}$:
$(-\hat{i} + \hat{j}) \times \hat{i} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & 0 \\ 1 & 0 & 0 \end{vmatrix} = \hat{k}(0-1) = -\hat{k}$.
We are given $(\overrightarrow{a} \times \overrightarrow{b}) \times \overrightarrow{c} = \lambda \overrightarrow{a} + \mu \overrightarrow{b}$.
Substituting the vectors: $-\hat{k} = \lambda(\hat{i} + \hat{j} + \hat{k}) + \mu(\hat{i} + \hat{j})$.
$-\hat{k} = (\lambda + \mu)\hat{i} + (\lambda + \mu)\hat{j} + \lambda\hat{k}$.
Comparing the coefficients on both sides:
For $\hat{i}$: $\lambda + \mu = 0$.
For $\hat{j}$: $\lambda + \mu = 0$.
For $\hat{k}$: $\lambda = -1$.
Since $\lambda + \mu = 0$,the value of $\lambda + \mu$ is $0$.

(B) The vector area of a triangle with vertices having position vectors $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ is given by the formula:
$\text{Vector Area} = \frac{1}{2} (\overrightarrow{AB} \times \overrightarrow{AC})$
Since $\overrightarrow{AB} = \overrightarrow{b} - \overrightarrow{a}$ and $\overrightarrow{AC} = \overrightarrow{c} - \overrightarrow{a}$,we have:
$\text{Vector Area} = \frac{1}{2} ((\overrightarrow{b} - \overrightarrow{a}) \times (\overrightarrow{c} - \overrightarrow{a}))$
Expanding the cross product:
$\text{Vector Area} = \frac{1}{2} (\overrightarrow{b} \times \overrightarrow{c} - \overrightarrow{b} \times \overrightarrow{a} - \overrightarrow{a} \times \overrightarrow{c} + \overrightarrow{a} \times \overrightarrow{a})$
Since $\overrightarrow{a} \times \overrightarrow{a} = 0$,$-\overrightarrow{b} \times \overrightarrow{a} = \overrightarrow{a} \times \overrightarrow{b}$,and $-\overrightarrow{a} \times \overrightarrow{c} = \overrightarrow{c} \times \overrightarrow{a}$,we get:
$\text{Vector Area} = \frac{1}{2} (\overrightarrow{a} \times \overrightarrow{b} + \overrightarrow{b} \times \overrightarrow{c} + \overrightarrow{c} \times \overrightarrow{a})$

(A) First,calculate the cross product $\vec{v} = \vec{b} \times \vec{c}$:
$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 2 \\ 1 & 2 & -1 \end{vmatrix} = \hat{i}(1-4) - \hat{j}(-1-2) + \hat{k}(2+1) = -3\hat{i} + 3\hat{j} + 3\hat{k}$
The magnitude is $|\vec{v}| = \sqrt{(-3)^2 + 3^2 + 3^2} = \sqrt{9+9+9} = \sqrt{27} = 3\sqrt{3}$.
Let $\theta$ be the angle between $\vec{a}$ and $\vec{v}$. Given $\cos \theta = \sqrt{\frac{2}{3}}$,we find $\sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - \frac{2}{3}} = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}}$.
The magnitude of the cross product is $|\vec{a} \times \vec{v}| = |\vec{a}| |\vec{v}| \sin \theta$.
Since $|\vec{a}| = 1$,we have $|\vec{a} \times \vec{v}| = 1 \cdot (3\sqrt{3}) \cdot \frac{1}{\sqrt{3}} = 3$.

(C) Given,$a=2 \hat{i}-2 \hat{j}+\hat{k}$,then $|a|=\sqrt{2^2+(-2)^2+1^2}=\sqrt{4+4+1}=3$.
Given $|c-a|=2 \sqrt{2}$. Squaring both sides,we get $|c-a|^2=8$.
$|c|^2+|a|^2-2(a \cdot c)=8$.
Substituting $|a|=3$ and $a \cdot c=|c|$,we have $|c|^2+9-2|c|=8$.
$|c|^2-2|c|+1=0 \Rightarrow (|c|-1)^2=0 \Rightarrow |c|=1$.
Now,calculate $a \times b$:
$a \times b = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -2 & 1 \\ 0 & -1 & 1 \end{vmatrix} = \hat{i}(-2+1) - \hat{j}(2-0) + \hat{k}(-2-0) = -\hat{i}-2 \hat{j}-2 \hat{k}$.
Then $|a \times b| = \sqrt{(-1)^2+(-2)^2+(-2)^2} = \sqrt{1+4+4} = 3$.
The magnitude of the cross product is given by $|(a \times b) \times c| = |a \times b| |c| \sin \theta$,where $\theta = \frac{\pi}{3}$.
$|(a \times b) \times c| = 3 \times 1 \times \sin \frac{\pi}{3} = 3 \times \frac{\sqrt{3}}{2} = \frac{3 \sqrt{3}}{2}$.

(A) Given that,$|x|=|y|=|z|=\sqrt{2}$ and $\theta=60^{\circ}$.
Thus,$x \cdot y = |x||y| \cos 60^{\circ} = (\sqrt{2})(\sqrt{2}) \times \frac{1}{2} = 1$.
Similarly,$y \cdot z = z \cdot x = 1$ and $x \cdot x = y \cdot y = z \cdot z = |x|^2 = 2$.
Now,$a = x \times (y \times z) = (x \cdot z)y - (x \cdot y)z = 1 \cdot y - 1 \cdot z = y - z$ $\dots(i)$.
Also,$b = y \times (z \times x) = (y \cdot x)z - (y \cdot z)x = 1 \cdot z - 1 \cdot x = z - x$ $\dots(ii)$.
Adding $(i)$ and $(ii)$,we get $a + b = y - x$,so $y - x = a + b$ $\dots(iii)$.
Given $c = x \times y$. Taking cross product with $x$ and $y$ respectively:
$x \times c = x \times (x \times y) = (x \cdot y)x - (x \cdot x)y = x - 2y$ $\dots(iv)$.
$y \times c = y \times (x \times y) = (y \cdot y)x - (y \cdot x)y = 2x - y$ $\dots(v)$.
Subtracting $(v)$ from $(iv)$,we get $(x - y) \times c = (x - 2y) - (2x - y) = -x - y$,which implies $x + y = (y - x) \times c$.
Substituting $y - x = a + b$ from $(iii)$,we get $x + y = (a + b) \times c$ $\dots(vi)$.
Subtracting $(iii)$ from $(vi)$,we get $(x + y) - (y - x) = (a + b) \times c - (a + b)$.
$2x = (a + b) \times c - (a + b)$.
Therefore,$x = \frac{1}{2}[(a + b) \times c - (a + b)]$.

(D) Given,$\vec{r} \times \vec{a}=\vec{b}$.
Taking the cross product with $\vec{a}$ on both sides: $(\vec{r} \times \vec{a}) \times \vec{a}=\vec{b} \times \vec{a}$.
Using the vector triple product formula $(\vec{A} \times \vec{B}) \times \vec{C} = (\vec{A} \cdot \vec{C})\vec{B} - (\vec{B} \cdot \vec{C})\vec{A}$,we get:
$(\vec{r} \cdot \vec{a})\vec{a} - (\vec{a} \cdot \vec{a})\vec{r} = \vec{b} \times \vec{a}$.
Since $|\vec{a}|=2$,we have $\vec{a} \cdot \vec{a} = |\vec{a}|^2 = 4$.
Thus,$4\vec{r} = (\vec{r} \cdot \vec{a})\vec{a} - (\vec{b} \times \vec{a})$.
Now,$|\vec{r} \times \vec{a}| = |\vec{b}| \Rightarrow |\vec{r}||\vec{a}|\sin \theta = \sqrt{3}$.
$1 \times 2 \sin \theta = \sqrt{3} \Rightarrow \sin \theta = \frac{\sqrt{3}}{2} \Rightarrow \theta = \frac{\pi}{3}$.
Then,$\vec{r} \cdot \vec{a} = |\vec{r}||\vec{a}|\cos \theta = 1 \times 2 \times \cos \frac{\pi}{3} = 2 \times \frac{1}{2} = 1$.
Substituting $\vec{r} \cdot \vec{a} = 1$ into the equation $4\vec{r} = (\vec{r} \cdot \vec{a})\vec{a} - (\vec{b} \times \vec{a})$:
$4\vec{r} = \vec{a} - (\vec{b} \times \vec{a}) \Rightarrow \vec{r} = \frac{1}{4}[\vec{a} - (\vec{b} \times \vec{a})]$.

(D) Given $\bar{a} = \bar{i} - 2\bar{j} - 2\bar{k}$ and $\bar{b} = 2\bar{i} + \bar{j} + 2\bar{k}$.
We need to compute $(\bar{a} + 2\bar{b}) \times (3\bar{a} - \bar{b})$.
Using the distributive property of the cross product:
$(\bar{a} + 2\bar{b}) \times (3\bar{a} - \bar{b}) = \bar{a} \times (3\bar{a}) - \bar{a} \times \bar{b} + (2\bar{b}) \times (3\bar{a}) - (2\bar{b}) \times \bar{b}$.
Since $\bar{v} \times \bar{v} = 0$,we have $\bar{a} \times \bar{a} = 0$ and $\bar{b} \times \bar{b} = 0$.
So,the expression simplifies to: $0 - (\bar{a} \times \bar{b}) + 6(\bar{b} \times \bar{a}) - 0$.
Since $\bar{b} \times \bar{a} = -(\bar{a} \times \bar{b})$,we get:
$-(\bar{a} \times \bar{b}) - 6(\bar{a} \times \bar{b}) = -7(\bar{a} \times \bar{b})$.
Now,calculate $\bar{a} \times \bar{b} = \begin{vmatrix} \bar{i} & \bar{j} & \bar{k} \\ 1 & -2 & -2 \\ 2 & 1 & 2 \end{vmatrix} = \bar{i}(-4 - (-2)) - \bar{j}(2 - (-4)) + \bar{k}(1 - (-4)) = -2\bar{i} - 6\bar{j} + 5\bar{k}$.
Finally,$-7(\bar{a} \times \bar{b}) = -7(-2\bar{i} - 6\bar{j} + 5\bar{k}) = 14\bar{i} + 42\bar{j} - 35\bar{k}$.

(B) Since $\bar{d}$ is perpendicular to both $\bar{a}$ and $\bar{b}$,$\bar{d}$ must be parallel to $\bar{a} \times \bar{b}$.
$\bar{a} \times \bar{b} = \begin{vmatrix} \bar{i} & \bar{j} & \bar{k} \\ 1 & -1 & 1 \\ 1 & -2 & -2 \end{vmatrix} = \bar{i}(2+2) - \bar{j}(-2-1) + \bar{k}(-2+1) = 4\bar{i} + 3\bar{j} - \bar{k}$.
Let $\bar{d} = k(4\bar{i} + 3\bar{j} - \bar{k})$ for some scalar $k$.
Given $|\bar{d} \times \bar{c}| = 14$. Note that $\bar{d} \times \bar{c} = k(4\bar{i} + 3\bar{j} - \bar{k}) \times (6\bar{i} + 3\bar{j} - 2\bar{k})$.
$(4\bar{i} + 3\bar{j} - \bar{k}) \times (6\bar{i} + 3\bar{j} - 2\bar{k}) = \begin{vmatrix} \bar{i} & \bar{j} & \bar{k} \\ 4 & 3 & -1 \\ 6 & 3 & -2 \end{vmatrix} = \bar{i}(-6+3) - \bar{j}(-8+6) + \bar{k}(12-18) = -3\bar{i} + 2\bar{j} - 6\bar{k}$.
Magnitude is $\sqrt{(-3)^2 + 2^2 + (-6)^2} = \sqrt{9+4+36} = \sqrt{49} = 7$.
So,$|\bar{d} \times \bar{c}| = |k| \times 7 = 14$,which implies $|k| = 2$.
Now,$|\bar{d} \cdot \bar{c}| = |k(4\bar{i} + 3\bar{j} - \bar{k}) \cdot (6\bar{i} + 3\bar{j} - 2\bar{k})| = |k| |24 + 9 + 2| = 2 \times 35 = 70$.

(C) Let the point $P(\vec{r})$ be $x\hat{i}+y\hat{j}+z\hat{k}$. The fixed points are $A(1, 0, 0)$ and $B(0, 1, 0)$.
Vector $\vec{AP} = (x-1)\hat{i} + y\hat{j} + z\hat{k}$ and vector $\vec{AB} = -\hat{i} + \hat{j}$.
The area of $\triangle ABP$ is given by $\frac{1}{2} |\vec{AP} \times \vec{AB}|$.
Calculating the cross product:
$\vec{AP} \times \vec{AB} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ x-1 & y & z \\ -1 & 1 & 0 \end{vmatrix} = \hat{i}(0-z) - \hat{j}(0 - (-z)) + \hat{k}((x-1) - (-y)) = -z\hat{i} - z\hat{j} + (x+y-1)\hat{k}$.
The magnitude is $|\vec{AP} \times \vec{AB}| = \sqrt{(-z)^2 + (-z)^2 + (x+y-1)^2} = \sqrt{2z^2 + (x+y-1)^2}$.
Given the area is $1$,we have $1 = \frac{1}{2} \sqrt{2z^2 + (x+y-1)^2}$.
Squaring both sides: $2 = \sqrt{2z^2 + (x+y-1)^2} \Rightarrow 4 = 2z^2 + (x+y-1)^2$.