Scalar triple product and their applications Questions in English

(D) Three vectors $\vec{u}, \vec{v}, \vec{w}$ are coplanar if and only if their scalar triple product is zero,i.e.,$[\vec{u}, \vec{v}, \vec{w}] = 0$.
Let $\vec{u} = \vec{a} + \lambda \vec{b} + 3\vec{c}$,$\vec{v} = -2\vec{a} + 3\vec{b} - 4\vec{c}$,and $\vec{w} = \vec{a} - 3\vec{b} + 5\vec{c}$.
The condition $[\vec{u}, \vec{v}, \vec{w}] = 0$ implies the determinant of the coefficients of $\vec{a}, \vec{b}, \vec{c}$ must be zero:
$\begin{vmatrix} 1 & \lambda & 3 \\ -2 & 3 & -4 \\ 1 & -3 & 5 \end{vmatrix} = 0$
Expanding the determinant along the first row:
$1(3 \times 5 - (-4) \times (-3)) - \lambda((-2) \times 5 - (-4) \times 1) + 3((-2) \times (-3) - 3 \times 1) = 0$
$1(15 - 12) - \lambda(-10 + 4) + 3(6 - 3) = 0$
$1(3) - \lambda(-6) + 3(3) = 0$
$3 + 6\lambda + 9 = 0$
$6\lambda + 12 = 0$
$6\lambda = -12$
$\lambda = -2$.

(B) Given $d = \lambda a + \mu b + \nu c$.
To find $\lambda$,we take the scalar triple product of $d$ with $b$ and $c$:
$d \cdot (b \times c) = (\lambda a + \mu b + \nu c) \cdot (b \times c)$.
Since $b \cdot (b \times c) = 0$ and $c \cdot (b \times c) = 0$,the expression simplifies to:
$d \cdot (b \times c) = \lambda [a, b, c]$.
Therefore,$\lambda = \frac{[d, b, c]}{[a, b, c]}$.
Using the cyclic property of the scalar triple product,$[d, b, c] = [b, c, d]$ and $[a, b, c] = [b, c, a]$.
Thus,$\lambda = \frac{[b, c, d]}{[b, c, a]}$.

(C) Let the expression be $E = (\bar{u} + \bar{v} - \bar{w}) \cdot ((\bar{u} - \bar{v}) \times (\bar{v} - \bar{w}))$.
First,compute the cross product: $(\bar{u} - \bar{v}) \times (\bar{v} - \bar{w}) = \bar{u} \times \bar{v} - \bar{u} \times \bar{w} - \bar{v} \times \bar{v} + \bar{v} \times \bar{w}$.
Since $\bar{v} \times \bar{v} = 0$,this simplifies to $\bar{u} \times \bar{v} + \bar{w} \times \bar{u} + \bar{v} \times \bar{w}$.
Now,take the dot product with $(\bar{u} + \bar{v} - \bar{w})$:
$E = (\bar{u} + \bar{v} - \bar{w}) \cdot (\bar{u} \times \bar{v} + \bar{w} \times \bar{u} + \bar{v} \times \bar{w})$.
Distributing the dot product:
$= \bar{u} \cdot (\bar{u} \times \bar{v}) + \bar{u} \cdot (\bar{w} \times \bar{u}) + \bar{u} \cdot (\bar{v} \times \bar{w}) + \bar{v} \cdot (\bar{u} \times \bar{v}) + \bar{v} \cdot (\bar{w} \times \bar{u}) + \bar{v} \cdot (\bar{v} \times \bar{w}) - \bar{w} \cdot (\bar{u} \times \bar{v}) - \bar{w} \cdot (\bar{w} \times \bar{u}) - \bar{w} \cdot (\bar{v} \times \bar{w})$.
Using the property that scalar triple product is zero if any two vectors are the same:
$= 0 + 0 + [\bar{u} \bar{v} \bar{w}] + 0 + [\bar{v} \bar{w} \bar{u}] + 0 - [\bar{w} \bar{u} \bar{v}] - 0 - 0$.
Since $[\bar{u} \bar{v} \bar{w}] = [\bar{v} \bar{w} \bar{u}] = [\bar{w} \bar{u} \bar{v}]$,we get:
$E = [\bar{u} \bar{v} \bar{w}] + [\bar{u} \bar{v} \bar{w}] - [\bar{u} \bar{v} \bar{w}] = [\bar{u} \bar{v} \bar{w}] = \bar{u} \cdot (\bar{v} \times \bar{w})$.

(A) Since $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar,they form a basis for the space of vectors. Any vector $\vec{r}$ can be expressed as a linear combination of $\vec{a}, \vec{b}, \vec{c}$:
$\vec{r} = x_1 \vec{a} + x_2 \vec{b} + x_3 \vec{c}$
Taking the scalar triple product with $\vec{b}$ and $\vec{c}$:
$\vec{r} \cdot (\vec{b} \times \vec{c}) = x_1 \vec{a} \cdot (\vec{b} \times \vec{c}) + x_2 \vec{b} \cdot (\vec{b} \times \vec{c}) + x_3 \vec{c} \cdot (\vec{b} \times \vec{c})$
Since $\vec{b} \cdot (\vec{b} \times \vec{c}) = 0$ and $\vec{c} \cdot (\vec{b} \times \vec{c}) = 0$,we get:
$[\vec{r} \, \vec{b} \, \vec{c}] = x_1 [\vec{a} \, \vec{b} \, \vec{c}] \implies x_1 = \frac{[\vec{r} \, \vec{b} \, \vec{c}]}{[\vec{a} \, \vec{b} \, \vec{c}]} = \frac{[\vec{b} \, \vec{c} \, \vec{r}]}{[\vec{a} \, \vec{b} \, \vec{c}]}$
Similarly,$x_2 = \frac{[\vec{c} \, \vec{a} \, \vec{r}]}{[\vec{a} \, \vec{b} \, \vec{c}]}$ and $x_3 = \frac{[\vec{a} \, \vec{b} \, \vec{r}]}{[\vec{a} \, \vec{b} \, \vec{c}]}$
Substituting these into the expression for $\vec{r}$:
$\vec{r} = \frac{[\vec{b} \, \vec{c} \, \vec{r}]}{[\vec{a} \, \vec{b} \, \vec{c}]} \vec{a} + \frac{[\vec{c} \, \vec{a} \, \vec{r}]}{[\vec{a} \, \vec{b} \, \vec{c}]} \vec{b} + \frac{[\vec{a} \, \vec{b} \, \vec{r}]}{[\vec{a} \, \vec{b} \, \vec{c}]} \vec{c}$
Multiplying by $[\vec{a} \, \vec{b} \, \vec{c}]$:
$[\vec{b} \, \vec{c} \, \vec{r}] \vec{a} + [\vec{c} \, \vec{a} \, \vec{r}] \vec{b} + [\vec{a} \, \vec{b} \, \vec{r}] \vec{c} = [\vec{a} \, \vec{b} \, \vec{c}] \vec{r}$

(D) Since the vector $\overline{c}$ lies in the plane of $\overline{a}$ and $\overline{b}$,the scalar triple product $[\overline{a} \, \overline{b} \, \overline{c}]$ must be equal to $0$.
This is given by the determinant:
$\begin{vmatrix} 1 & 1 & 1 \\ 1 & -1 & 2 \\ x & x-2 & -1 \end{vmatrix} = 0$
Expanding the determinant along the first row:
$1((-1)(-1) - (2)(x-2)) - 1((1)(-1) - (2)(x)) + 1((1)(x-2) - (-1)(x)) = 0$
$1(1 - 2x + 4) - 1(-1 - 2x) + 1(x - 2 + x) = 0$
$(5 - 2x) + (1 + 2x) + (2x - 2) = 0$
$5 - 2x + 1 + 2x + 2x - 2 = 0$
$2x + 4 = 0$
$2x = -4$
$x = -2$

(C) The scalar triple product $a \cdot (b \times c)$ is given by the determinant of the matrix formed by the components of vectors $a$,$b$,and $c$:
$a \cdot (b \times c) = \begin{vmatrix} 2 & 1 & -1 \\ 1 & 2 & 1 \\ 1 & -1 & 2 \end{vmatrix}$
Expanding the determinant along the first row:
$= 2(2(2) - 1(-1)) - 1(1(2) - 1(1)) + (-1)(1(-1) - 2(1))$
$= 2(4 + 1) - 1(2 - 1) - 1(-1 - 2)$
$= 2(5) - 1(1) - 1(-3)$
$= 10 - 1 + 3$
$= 12$

(D) Three vectors $\vec{a}$,$\vec{b}$,and $\vec{c}$ are coplanar if and only if their scalar triple product is zero,i.e.,$[\vec{a} \, \vec{b} \, \vec{c}] = 0$.
This is equivalent to the determinant of the matrix formed by their components being zero:
$\begin{vmatrix} 1 & -2 & 3 \\ 2 & 3 & -1 \\ \lambda & 1 & 2\lambda - 1 \end{vmatrix} = 0$
Expanding the determinant along the first row:
$1[3(2\lambda - 1) - (-1)(1)] - (-2)[2(2\lambda - 1) - (-1)(\lambda)] + 3[2(1) - 3(\lambda)] = 0$
$1[6\lambda - 3 + 1] + 2[4\lambda - 2 + \lambda] + 3[2 - 3\lambda] = 0$
$(6\lambda - 2) + 2(5\lambda - 2) + (6 - 9\lambda) = 0$
$6\lambda - 2 + 10\lambda - 4 + 6 - 9\lambda = 0$
$(6 + 10 - 9)\lambda + (-2 - 4 + 6) = 0$
$7\lambda + 0 = 0$
$7\lambda = 0$
$\lambda = 0$
Thus,the correct option is $D$.

(A) Let the scalar triple product be denoted as $[\vec{a}, \vec{b}, \vec{c}] = \vec{a} \cdot (\vec{b} \times \vec{c})$.
We know that the scalar triple product is cyclic,so $[\vec{a}, \vec{b}, \vec{c}] = [\vec{b}, \vec{c}, \vec{a}] = [\vec{c}, \vec{a}, \vec{b}]$.
Also,swapping any two vectors changes the sign: $[\vec{a}, \vec{c}, \vec{b}] = -[\vec{a}, \vec{b}, \vec{c}]$.
Given expression: $E = \frac{[\vec{a}, \vec{b}, \vec{c}]}{[\vec{c}, \vec{a}, \vec{b}]} + \frac{[\vec{b}, \vec{a}, \vec{c}]}{[\vec{c}, \vec{a}, \vec{b}]}$.
Since $[\vec{c}, \vec{a}, \vec{b}] = [\vec{a}, \vec{b}, \vec{c}]$ and $[\vec{b}, \vec{a}, \vec{c}] = -[\vec{a}, \vec{b}, \vec{c}]$,
$E = \frac{[\vec{a}, \vec{b}, \vec{c}]}{[\vec{a}, \vec{b}, \vec{c}]} + \frac{-[\vec{a}, \vec{b}, \vec{c}]}{[\vec{a}, \vec{b}, \vec{c}]}$.
$E = 1 - 1 = 0$.

(D) The scalar triple product is given by the determinant of the components of the vectors $\vec{a}, \vec{b},$ and $\vec{c}$.
$[\vec{a} \, \vec{b} \, \vec{c}] = \begin{vmatrix} 1 & 0 & -1 \\ x & 1 & 1 - x \\ y & x & 1 + x - y \end{vmatrix}$
Applying the column operation $C_3 \to C_3 + C_1$:
$[\vec{a} \, \vec{b} \, \vec{c}] = \begin{vmatrix} 1 & 0 & 0 \\ x & 1 & 1 \\ y & x & 1 + x \end{vmatrix}$
Expanding along the first row:
$[\vec{a} \, \vec{b} \, \vec{c}] = 1 \times \begin{vmatrix} 1 & 1 \\ x & 1 + x \end{vmatrix} - 0 + 0$
$[\vec{a} \, \vec{b} \, \vec{c}] = (1 + x) - x = 1$
Since the result is a constant $1$,it does not depend on $x$ or $y$.

(D) Let the four points be $A, B, C,$ and $D$ represented by position vectors $\vec{p_1} = 2\vec{a} + 3\vec{b} - \vec{c}$,$\vec{p_2} = \vec{a} - 2\vec{b} + 3\vec{c}$,$\vec{p_3} = 3\vec{a} + 4\vec{b} - 2\vec{c}$,and $\vec{p_4} = \vec{a} - \lambda\vec{b} - 6\vec{c}$.
For the points to be coplanar,the vectors $\vec{AB}, \vec{AC},$ and $\vec{AD}$ must be coplanar,meaning their scalar triple product is zero: $[\vec{AB}, \vec{AC}, \vec{AD}] = 0$.
$\vec{AB} = \vec{p_2} - \vec{p_1} = -\vec{a} - 5\vec{b} + 4\vec{c}$
$\vec{AC} = \vec{p_3} - \vec{p_1} = \vec{a} + \vec{b} - \vec{c}$
$\vec{AD} = \vec{p_4} - \vec{p_1} = -\vec{a} - (\lambda + 3)\vec{b} - 5\vec{c}$
The scalar triple product is given by the determinant:
$\begin{vmatrix} -1 & -5 & 4 \\ 1 & 1 & -1 \\ -1 & -(\lambda + 3) & -5 \end{vmatrix} = 0$
Expanding the determinant:
$-1(-5 - (\lambda + 3)) + 5(-5 - 1) + 4(-(\lambda + 3) + 1) = 0$
$-1(-5 - \lambda - 3) + 5(-6) + 4(-\lambda - 3 + 1) = 0$
$-1(-\lambda - 8) - 30 + 4(-\lambda - 2) = 0$
$\lambda + 8 - 30 - 4\lambda - 8 = 0$
$-3\lambda - 30 = 0$
$-3\lambda = 30 \implies \lambda = -10$.
Since $-10$ is not among the options,the correct answer is $D$.

(A) The scalar triple product is defined as $[x, y, z] = x \cdot (y \times z)$.
Given that $a, b,$ and $c$ are coplanar vectors,their scalar triple product $[a, b, c] = a \cdot (b \times c) = 0$.
We need to evaluate $[2a - b, 2b - c, 2c - a]$.
Using the properties of the scalar triple product,we can expand this as:
$[2a - b, 2b - c, 2c - a] = (2a - b) \cdot ((2b - c) \times (2c - a))$.
First,calculate the cross product $(2b - c) \times (2c - a)$:
$(2b - c) \times (2c - a) = 4(b \times c) - 2(b \times a) - 2(c \times c) + (c \times a)$.
Since $c \times c = 0$,this simplifies to $4(b \times c) + 2(a \times b) + (c \times a)$.
Now,take the dot product with $(2a - b)$:
$(2a - b) \cdot [4(b \times c) + 2(a \times b) + (c \times a)] = 8[a, b, c] + 4[a, a, b] + 2[a, c, a] - 4[b, b, c] - 2[b, a, b] - [b, c, a]$.
Since any scalar triple product with two identical vectors is $0$ (e.g.,$[a, a, b] = 0$),the expression simplifies to:
$8[a, b, c] - [b, c, a]$.
Since $[a, b, c] = [b, c, a]$,the expression becomes $8[a, b, c] - [a, b, c] = 7[a, b, c]$.
Given that $a, b,$ and $c$ are coplanar,$[a, b, c] = 0$.
Therefore,$7 \times 0 = 0$.

(A) Three vectors $\vec{u}, \vec{v}, \vec{w}$ are coplanar if their scalar triple product is zero,i.e.,$[\vec{u}, \vec{v}, \vec{w}] = 0$.
Given vectors are $\vec{u} = a\hat{i} + a\hat{j} + c\hat{k}$,$\vec{v} = \hat{i} + 0\hat{j} + \hat{k}$,and $\vec{w} = c\hat{i} + c\hat{j} + b\hat{k}$.
The scalar triple product is given by the determinant:
$\begin{vmatrix} a & a & c \\ 1 & 0 & 1 \\ c & c & b \end{vmatrix} = 0$
Expanding along the second row:
$-1(ab - c^2) + 0 - 1(ac - ac) = 0$
$-(ab - c^2) = 0$
$c^2 - ab = 0$
$c^2 = ab$
$c = \sqrt{ab}$
Thus,$c$ is the geometric mean of $a$ and $b$.

(D) For three vectors to be coplanar,their scalar triple product must be zero,i.e.,$[\vec{a}, \vec{b}, \vec{c}] = 0$.
The vectors are $\vec{a} = 1i + 3j + 0k$,$\vec{b} = 0i + 0j + 5k$,and $\vec{c} = Pi - 1j + 0k$.
The condition for coplanarity is given by the determinant:
$\begin{vmatrix} 1 & 3 & 0 \\ 0 & 0 & 5 \\ P & -1 & 0 \end{vmatrix} = 0$
Expanding along the first row:
$1(0 - (-5)) - 3(0 - 5P) + 0(0 - 0) = 0$
$1(5) - 3(-5P) = 0$
$5 + 15P = 0$
$15P = -5$
$P = -5/15 = -1/3$
Thus,the value of $P$ is $-1/3$.

(A) Given $\bar{a} \cdot \bar{d} = 0$,this implies $\bar{d} \perp \bar{a}$.
Given $[\bar{b} \bar{c} \bar{d}] = 0$,this implies $\bar{b}, \bar{c}$,and $\bar{d}$ are coplanar,which means $\bar{d} \perp (\bar{b} \times \bar{c})$.
Therefore,$\bar{d}$ is parallel to $\bar{a} \times (\bar{b} \times \bar{c})$.
Since $\bar{d}$ is a unit vector,$\bar{d} = \pm \frac{\bar{a} \times (\bar{b} \times \bar{c})}{|\bar{a} \times (\bar{b} \times \bar{c})|}$.
Here,$\bar{a} = (1, -1, 0)$,$\bar{b} = (0, 1, -1)$,and $\bar{c} = (-1, 0, 1)$.
Calculating the cross product $\bar{b} \times \bar{c} = \begin{vmatrix} \bar{i} & \bar{j} & \bar{k} \\ 0 & 1 & -1 \\ -1 & 0 & 1 \end{vmatrix} = \bar{i}(1-0) - \bar{j}(0-1) + \bar{k}(0+1) = (1, 1, 1)$.
Now,$\bar{a} \times (\bar{b} \times \bar{c}) = \begin{vmatrix} \bar{i} & \bar{j} & \bar{k} \\ 1 & -1 & 0 \\ 1 & 1 & 1 \end{vmatrix} = \bar{i}(-1-0) - \bar{j}(1-0) + \bar{k}(1+1) = (-1, -1, 2)$.
The magnitude is $|\bar{a} \times (\bar{b} \times \bar{c})| = \sqrt{(-1)^2 + (-1)^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$.
Thus,$\bar{d} = \pm \frac{1}{\sqrt{6}} (-\bar{i} - \bar{j} + 2\bar{k}) = \pm \frac{1}{\sqrt{6}} (\bar{i} + \bar{j} - 2\bar{k})$.

(B) Statement-$1$: The scalar triple product $\vec{a} \cdot (\vec{b} \times \vec{c})$ represents the volume of the parallelepiped formed by vectors $\vec{a}, \vec{b},$ and $\vec{c}$. If the volume is $0$,the vectors must lie in the same plane (coplanar). Thus,Statement-$1$ is true.
Statement-$2$: Two non-zero vectors are perpendicular if their dot product is $0$. The cross product $\vec{u} \times \vec{v}$ is indeed a vector perpendicular to the plane containing $\vec{u}$ and $\vec{v}$. Thus,Statement-$2$ is true.
However,Statement-$2$ provides the condition for the perpendicularity of two vectors,which is a fundamental property of the dot product,but it does not explain why the scalar triple product of three vectors is zero when they are coplanar. Therefore,Statement-$2$ is not the correct explanation for Statement-$1$.

(B) The position vectors of the vertices $A, B, C, D$ are $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ respectively.
$\vec{AB} = \vec{b} - \vec{a} = 3\vec{i} - \vec{j} - \vec{k}$
$\vec{AC} = \vec{c} - \vec{a} = 4\vec{i} + 2\vec{j} + 4\vec{k}$
$\vec{AD} = \vec{d} - \vec{a} = 2\vec{i} + 2\vec{j}$
The volume of the tetrahedron is given by $V = \frac{1}{6} |[\vec{AB}, \vec{AC}, \vec{AD}]|$.
$V = \frac{1}{6} \left| \det \begin{bmatrix} 3 & -1 & -1 \\ 4 & 2 & 4 \\ 2 & 2 & 0 \end{bmatrix} \right|$
Calculating the determinant: $3(0 - 8) - (-1)(0 - 8) + (-1)(8 - 4) = 3(-8) + 1(-8) - 1(4) = -24 - 8 - 4 = -36$.
Taking the absolute value: $V = \frac{1}{6} |-36| = \frac{36}{6} = 6$ cubic units.

(B) Let $V = [\vec{a} \vec{b} \vec{c}] = \vec{a} \cdot (\vec{b} \times \vec{c})$.
Then $\vec{p} = \frac{\vec{b} \times \vec{c}}{V}, \vec{q} = \frac{\vec{c} \times \vec{a}}{V}, \vec{r} = \frac{\vec{a} \times \vec{b}}{V}$.
$[\vec{p} \vec{q} \vec{r}] = \vec{p} \cdot (\vec{q} \times \vec{r}) = \frac{\vec{b} \times \vec{c}}{V} \cdot \left( \frac{\vec{c} \times \vec{a}}{V} \times \frac{\vec{a} \times \vec{b}}{V} \right)$.
Using the vector triple product identity $(\vec{u} \times \vec{v}) \times \vec{w} = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{v} \cdot \vec{w})\vec{u}$,we have $(\vec{c} \times \vec{a}) \times (\vec{a} \times \vec{b}) = [(\vec{c} \times \vec{a}) \cdot \vec{b}]\vec{a} - [(\vec{c} \times \vec{a}) \cdot \vec{a}]\vec{b}$.
Since $(\vec{c} \times \vec{a}) \cdot \vec{a} = 0$,this simplifies to $[\vec{c} \vec{a} \vec{b}]\vec{a} = [\vec{a} \vec{b} \vec{c}]\vec{a} = V\vec{a}$.
Thus,$[\vec{p} \vec{q} \vec{r}] = \frac{1}{V^3} (\vec{b} \times \vec{c}) \cdot (V\vec{a}) = \frac{V}{V^3} (\vec{b} \times \vec{c}) \cdot \vec{a} = \frac{V^2}{V^3} = \frac{1}{V} = \frac{1}{[\vec{a} \vec{b} \vec{c}]}$.

(D) By the definition of the Scalar Triple Product,$[\vec{a} \, \vec{b} \, \vec{c}] = \vec{a} \cdot (\vec{b} \times \vec{c})$.
When $[\vec{a} \, \vec{b} \, \vec{c}] = 0$,it implies that the volume of the parallelepiped formed by the vectors $\vec{a}, \vec{b},$ and $\vec{c}$ is zero.
This condition is only possible if the three vectors lie in the same plane,i.e.,they are coplanar.
Since the definition $[\vec{a} \, \vec{b} \, \vec{c}] = \vec{a} \cdot (\vec{b} \times \vec{c})$ is equivalent to the condition of being coplanar,both $B$ and $C$ are mathematically correct statements. However,in standard vector algebra,the condition $[\vec{a} \, \vec{b} \, \vec{c}] = 0$ is the defining condition for vectors to be coplanar. Given the options,$B$ is the geometric interpretation,while $C$ is the algebraic definition. Since $C$ is the definition of the expression itself,and $B$ is the consequence,if multiple choices are allowed,$D$ would be the most comprehensive answer. Assuming a single correct choice,$B$ is the standard geometric result.

(C) The volume $V$ of a parallelepiped with coterminous edges $\vec{a}, \vec{b}, \vec{c}$ is given by the absolute value of the scalar triple product: $V = |[\vec{a} \vec{b} \vec{c}]|$.
Calculating the determinant:
$V = \left| \det \begin{bmatrix} 1 & a & 1 \\ 0 & 1 & a \\ a & 0 & 1 \end{bmatrix} \right| = |1(1-0) - a(0-a^2) + 1(0-a)| = |1 + a^3 - a|$.
To find the minimum volume,we analyze $f(a) = a^3 - a + 1$.
Taking the derivative with respect to $a$: $f'(a) = 3a^2 - 1$.
Setting $f'(a) = 0$,we get $3a^2 = 1$,which implies $a = \pm \frac{1}{\sqrt{3}}$.
Using the second derivative test: $f''(a) = 6a$.
For $a = \frac{1}{\sqrt{3}}$,$f''(\frac{1}{\sqrt{3}}) = 6(\frac{1}{\sqrt{3}}) = 2\sqrt{3} > 0$,which indicates a local minimum.
Thus,the volume is minimum at $a = \frac{1}{\sqrt{3}}$.

(A) We know that $[\bar{a} \times \bar{b}, \bar{b} \times \bar{c}, \bar{c} \times \bar{a}] = (\bar{a} \times \bar{b}) \cdot [(\bar{b} \times \bar{c}) \times (\bar{c} \times \bar{a})].$
Using the vector triple product formula $(\vec{x} \times \vec{y}) \times \vec{z} = (\vec{x} \cdot \vec{z})\vec{y} - (\vec{y} \cdot \vec{z})\vec{x},$ we have:
$(\bar{b} \times \bar{c}) \times (\bar{c} \times \bar{a}) = [(\bar{b} \times \bar{c}) \cdot \bar{a}]\bar{c} - [(\bar{b} \times \bar{c}) \cdot \bar{c}]\bar{a}.$
Since $[(\bar{b} \times \bar{c}) \cdot \bar{c}] = 0,$ this simplifies to $[\bar{a} \bar{b} \bar{c}]\bar{c}.$
Thus,the expression becomes $(\bar{a} \times \bar{b}) \cdot ([\bar{a} \bar{b} \bar{c}]\bar{c}) = [\bar{a} \bar{b} \bar{c}] ((\bar{a} \times \bar{b}) \cdot \bar{c}) = [\bar{a} \bar{b} \bar{c}]^2.$
Given $[\bar{a} \bar{b} \bar{c}] = 4,$ the value is $4^2 = 16.$

(D) The scalar triple product is defined as $[\bar{a} - 2\bar{b}, \bar{b} - 3\bar{c}, \bar{c} - 4\bar{a}] = ((\bar{a} - 2\bar{b}) \times (\bar{b} - 3\bar{c})) \cdot (\bar{c} - 4\bar{a})$.
Expanding the cross product: $(\bar{a} \times \bar{b} - 3(\bar{a} \times \bar{c}) - 2(\bar{b} \times \bar{b}) + 6(\bar{b} \times \bar{c})) \cdot (\bar{c} - 4\bar{a})$.
Since $\bar{b} \times \bar{b} = 0$,this simplifies to $(\bar{a} \times \bar{b} - 3(\bar{a} \times \bar{c}) + 6(\bar{b} \times \bar{c})) \cdot (\bar{c} - 4\bar{a})$.
Distributing the dot product:
$= (\bar{a} \times \bar{b}) \cdot \bar{c} - 4(\bar{a} \times \bar{b}) \cdot \bar{a} - 3(\bar{a} \times \bar{c}) \cdot \bar{c} + 12(\bar{a} \times \bar{c}) \cdot \bar{a} + 6(\bar{b} \times \bar{c}) \cdot \bar{c} - 24(\bar{b} \times \bar{c}) \cdot \bar{a}$.
Since the scalar triple product is zero if any two vectors are the same,terms like $(\bar{a} \times \bar{b}) \cdot \bar{a}$ and $(\bar{a} \times \bar{c}) \cdot \bar{c}$ vanish.
$= [\bar{a}, \bar{b}, \bar{c}] - 24[\bar{b}, \bar{c}, \bar{a}] = [\bar{a}, \bar{b}, \bar{c}] - 24[\bar{a}, \bar{b}, \bar{c}] = -23[\bar{a}, \bar{b}, \bar{c}]$.
Given $\bar{a}, \bar{b}, \bar{c}$ are mutually perpendicular,$[\bar{a}, \bar{b}, \bar{c}] = |\bar{a}| |\bar{b}| |\bar{c}| = 1 \times 3 \times 5 = 15$.
Therefore,$-23 \times 15 = -345$.

(B) Since $\vec{a}$ and $\vec{b}$ are mutually perpendicular unit vectors,the set $\{\vec{a}, \vec{b}, \vec{a} \times \vec{b}\}$ forms an orthogonal basis for the space.
Let $\vec{r} = x_1 \vec{a} + x_2 \vec{b} + x_3 (\vec{a} \times \vec{b})$.
Taking the dot product with $\vec{a}$: $\vec{r} \cdot \vec{a} = x_1 |\vec{a}|^2 = x_1(1) = 0 \implies x_1 = 0$.
Taking the dot product with $\vec{b}$: $\vec{r} \cdot \vec{b} = x_2 |\vec{b}|^2 = x_2(1) = 1 \implies x_2 = 1$.
Taking the dot product with $(\vec{a} \times \vec{b})$: $\vec{r} \cdot (\vec{a} \times \vec{b}) = x_3 |\vec{a} \times \vec{b}|^2 = x_3(1)^2 = 1 \implies x_3 = 1$.
Substituting the values of $x_1, x_2, x_3$ into the expression for $\vec{r}$,we get $\vec{r} = 0\vec{a} + 1\vec{b} + 1(\vec{a} \times \vec{b}) = \vec{b} + (\vec{a} \times \vec{b})$.

(D) Three vectors $\vec{A}$,$\vec{B}$,and $\vec{C}$ are coplanar if and only if their scalar triple product is zero,i.e.,$[\vec{A} \, \vec{B} \, \vec{C}] = 0$.
This is equivalent to the determinant of the matrix formed by their components being zero:
$\begin{vmatrix} 2 & -1 & 1 \\ 1 & 2 & -3 \\ 3 & a & 5 \end{vmatrix} = 0$
Expanding the determinant along the first row:
$2(2 \times 5 - (-3) \times a) - (-1)(1 \times 5 - (-3) \times 3) + 1(1 \times a - 2 \times 3) = 0$
$2(10 + 3a) + 1(5 + 9) + (a - 6) = 0$
$20 + 6a + 14 + a - 6 = 0$
$7a + 28 = 0$
$7a = -28$
$a = -4$

(C) The volume of a parallelepiped with coterminous edges $\vec{OA}, \vec{OB},$ and $\vec{OC}$ is given by the absolute value of the scalar triple product $[\vec{OA}, \vec{OB}, \vec{OC}]$.
Given the coordinates $A(4, 3, 1), B(3, 1, 2),$ and $C(5, 2, 1)$,the vectors are $\vec{OA} = 4\hat{i} + 3\hat{j} + 1\hat{k}$,$\vec{OB} = 3\hat{i} + 1\hat{j} + 2\hat{k}$,and $\vec{OC} = 5\hat{i} + 2\hat{j} + 1\hat{k}$.
The scalar triple product is calculated as the determinant:
$[\vec{OA}, \vec{OB}, \vec{OC}] = \begin{vmatrix} 4 & 3 & 1 \\ 3 & 1 & 2 \\ 5 & 2 & 1 \end{vmatrix}$
Expanding the determinant along the first row:
$= 4(1(1) - 2(2)) - 3(3(1) - 5(2)) + 1(3(2) - 5(1))$
$= 4(1 - 4) - 3(3 - 10) + 1(6 - 5)$
$= 4(-3) - 3(-7) + 1(1)$
$= -12 + 21 + 1 = 10$
Since the volume must be positive,the volume is $|10| = 10$ cubic units.

(D) We know that the scalar triple product of the cross products of three vectors is given by the formula:
$\left[ {\vec a \times \vec b, \vec b \times \vec c, \vec c \times \vec a } \right] = \left[ {\vec a \,\vec b \,\vec c } \right]^2$.
Given that $\left[ {\vec a \,\vec b \,\vec c } \right] = 4$.
Substituting the value into the formula:
$\left[ {\vec a \times \vec b, \vec b \times \vec c, \vec c \times \vec a } \right] = (4)^2 = 16$.
Therefore,the correct option is $D$.

(B) We know that the cross products of unit vectors are: $\hat{i} \times \hat{j} = \hat{k}$,$\hat{j} \times \hat{k} = \hat{i}$,and $\hat{k} \times \hat{i} = \hat{j}$.
First,evaluate the expression inside the square brackets:
$(\hat{j} \times \hat{k}) \times (\hat{k} \times \hat{i}) = \hat{i} \times \hat{j} = \hat{k}$.
Now,substitute this back into the original expression:
$(\hat{i} \times \hat{j}) \cdot \hat{k} = \hat{k} \cdot \hat{k}$.
Since the dot product of a unit vector with itself is $1$,we have:
$\hat{k} \cdot \hat{k} = 1$.

(C) Given that $\vec{c}$ is a unit vector,$|\vec{c}| = 1$,so $c_1^2 + c_2^2 + c_3^2 = 1 \dots (i)$.
Since $\vec{c} \perp \vec{a}$ and $\vec{c} \perp \vec{b}$,we have $\vec{c} \cdot \vec{a} = 0 \Rightarrow a_1c_1 + a_2c_2 + a_3c_3 = 0 \dots (ii)$ and $\vec{c} \cdot \vec{b} = 0 \Rightarrow b_1c_1 + b_2c_2 + b_3c_3 = 0 \dots (iii)$.
The angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{6}$,so $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2} |\vec{a}| |\vec{b}|$.
Squaring both sides,$(\vec{a} \cdot \vec{b})^2 = \frac{3}{4} |\vec{a}|^2 |\vec{b}|^2 \dots (iv)$.
Now,let $D = \left| \begin{array}{ccc} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{array} \right|$. Then $D^2 = \left| \begin{array}{ccc} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{array} \right| \left| \begin{array}{ccc} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{array} \right|$.
Using the property of the determinant of the product of matrices,$D^2 = \left| \begin{array}{ccc} \vec{a} \cdot \vec{a} & \vec{a} \cdot \vec{b} & \vec{a} \cdot \vec{c} \\ \vec{b} \cdot \vec{a} & \vec{b} \cdot \vec{b} & \vec{b} \cdot \vec{c} \\ \vec{c} \cdot \vec{a} & \vec{c} \cdot \vec{b} & \vec{c} \cdot \vec{c} \end{array} \right| = \left| \begin{array}{ccc} |\vec{a}|^2 & \vec{a} \cdot \vec{b} & 0 \\ \vec{a} \cdot \vec{b} & |\vec{b}|^2 & 0 \\ 0 & 0 & 1 \end{array} \right|$.
Expanding along the third row,$D^2 = 1 \cdot (|\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2) = |\vec{a}|^2 |\vec{b}|^2 - \frac{3}{4} |\vec{a}|^2 |\vec{b}|^2 = \frac{1}{4} |\vec{a}|^2 |\vec{b}|^2$.

(B) Let the vertices be $A(\hat{i} - 6\hat{j} + 10\hat{k})$,$B(-\hat{i} - 3\hat{j} + 7\hat{k})$,$C(5\hat{i} - \hat{j} + \lambda\hat{k})$,and $D(7\hat{i} - 4\hat{j} + 7\hat{k})$.
The vectors representing the edges from vertex $A$ are:
$\vec{AB} = B - A = (-1-1)\hat{i} + (-3+6)\hat{j} + (7-10)\hat{k} = -2\hat{i} + 3\hat{j} - 3\hat{k}$
$\vec{AC} = C - A = (5-1)\hat{i} + (-1+6)\hat{j} + (\lambda-10)\hat{k} = 4\hat{i} + 5\hat{j} + (\lambda-10)\hat{k}$
$\vec{AD} = D - A = (7-1)\hat{i} + (-4+6)\hat{j} + (7-10)\hat{k} = 6\hat{i} + 2\hat{j} - 3\hat{k}$
The volume of a tetrahedron is given by $V = \frac{1}{6} |[\vec{AB} \vec{AC} \vec{AD}]| = 11$.
So,$|[\vec{AB} \vec{AC} \vec{AD}]| = 66$,which means $[\vec{AB} \vec{AC} \vec{AD}] = \pm 66$.
The scalar triple product is the determinant:
$\begin{vmatrix} -2 & 3 & -3 \\ 4 & 5 & \lambda-10 \\ 6 & 2 & -3 \end{vmatrix} = \pm 66$
Expanding the determinant:
$-2[5(-3) - 2(\lambda-10)] - 3[4(-3) - 6(\lambda-10)] - 3[4(2) - 6(5)] = \pm 66$
$-2[-15 - 2\lambda + 20] - 3[-12 - 6\lambda + 60] - 3[8 - 30] = \pm 66$
$-2[5 - 2\lambda] - 3[48 - 6\lambda] - 3[-22] = \pm 66$
$-10 + 4\lambda - 144 + 18\lambda + 66 = \pm 66$
$22\lambda - 88 = \pm 66$
Case $1$: $22\lambda - 88 = 66 \Rightarrow 22\lambda = 154 \Rightarrow \lambda = 7$
Case $2$: $22\lambda - 88 = -66 \Rightarrow 22\lambda = 22 \Rightarrow \lambda = 1$
Thus,$\lambda = 1, 7$.

(D) Let $\vec{a} = \hat{i} + \hat{j}$,$\vec{b} = \hat{j} + \hat{k}$,and $\vec{c} = \hat{k} + \hat{i}$.
The unit vectors perpendicular to the planes are given by:
$\vec{\alpha} = \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}$,$\vec{\beta} = \frac{\vec{b} \times \vec{c}}{|\vec{b} \times \vec{c}|}$,and $\vec{\gamma} = \frac{\vec{c} \times \vec{a}}{|\vec{c} \times \vec{a}|}$.
The volume of the parallelepiped formed by these vectors is the scalar triple product $|[\vec{\alpha}, \vec{\beta}, \vec{\gamma}]|$.
$|[\vec{\alpha}, \vec{\beta}, \vec{\gamma}]| = \frac{|[\vec{a} \times \vec{b}, \vec{b} \times \vec{c}, \vec{c} \times \vec{a}]|}{|\vec{a} \times \vec{b}| |\vec{b} \times \vec{c}| |\vec{c} \times \vec{a}|} = \frac{[\vec{a}, \vec{b}, \vec{c}]^2}{|\vec{a} \times \vec{b}| |\vec{b} \times \vec{c}| |\vec{c} \times \vec{a}|}$.
First,calculate the scalar triple product $[\vec{a}, \vec{b}, \vec{c}]$:
$[\vec{a}, \vec{b}, \vec{c}] = \begin{vmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{vmatrix} = 1(1-0) - 1(0-1) + 0 = 1 + 1 = 2$.
Next,calculate the cross products:
$\vec{a} \times \vec{b} = (\hat{i} + \hat{j}) \times (\hat{j} + \hat{k}) = \hat{k} - \hat{i} + \hat{j} = \hat{i} + \hat{j} + \hat{k}$.
$\vec{b} \times \vec{c} = (\hat{j} + \hat{k}) \times (\hat{k} + \hat{i}) = \hat{i} + \hat{j} + \hat{k}$.
$\vec{c} \times \vec{a} = (\hat{k} + \hat{i}) \times (\hat{i} + \hat{j}) = \hat{j} - \hat{k} + \hat{i} = \hat{i} + \hat{j} + \hat{k}$.
The magnitude of each cross product is $|\vec{a} \times \vec{b}| = |\vec{b} \times \vec{c}| = |\vec{c} \times \vec{a}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$.
Therefore,the volume is $\frac{2^2}{\sqrt{3} \cdot \sqrt{3} \cdot \sqrt{3}} = \frac{4}{3\sqrt{3}}$ cubic units.

(D) The volume $V$ of a parallelepiped with coterminous edges $\vec{u}, \vec{v}, \vec{w}$ is given by the scalar triple product $|[\vec{u}, \vec{v}, \vec{w}]| = |\vec{u} \cdot (\vec{v} \times \vec{w})|$.
Here,the edges are $\vec{u} = (a - b)$,$\vec{v} = (b - c)$,and $\vec{w} = (c - a)$.
Volume $V = |(a - b) \cdot ((b - c) \times (c - a))|$.
Expanding the cross product: $(b - c) \times (c - a) = (b \times c) - (b \times a) - (c \times c) + (c \times a)$.
Since $c \times c = 0$,we have $(b - c) \times (c - a) = (b \times c) - (b \times a) + (c \times a)$.
Now,calculate the dot product with $(a - b)$:
$V = (a - b) \cdot (b \times c - b \times a + c \times a)$
$V = a \cdot (b \times c) - a \cdot (b \times a) + a \cdot (c \times a) - b \cdot (b \times c) + b \cdot (b \times a) - b \cdot (c \times a)$.
Using the property of scalar triple products where any two vectors are identical,the result is $0$:
$a \cdot (b \times c) = [a, b, c]$
$a \cdot (b \times a) = 0$
$a \cdot (c \times a) = 0$
$b \cdot (b \times c) = 0$
$b \cdot (b \times a) = 0$
$b \cdot (c \times a) = [b, c, a] = [a, b, c]$.
Substituting these: $V = [a, b, c] - 0 + 0 - 0 + 0 - [a, b, c] = 0$.

(A) Let $\vec{d} = x\hat{i} + y\hat{j} + z\hat{k}$. Since $\vec{d}$ is a unit vector,$x^2 + y^2 + z^2 = 1 \dots (i)$.
Given $\vec{a} \cdot \vec{d} = 0$,we have $(\hat{i} - \hat{j}) \cdot (x\hat{i} + y\hat{j} + z\hat{k}) = 0$,which implies $x - y = 0$,so $x = y$.
Given $[\vec{b} \, \vec{c} \, \vec{d}] = 0$,the scalar triple product is zero,meaning $\vec{d}$ is coplanar with $\vec{b}$ and $\vec{c}$. Thus,$\vec{d} = \lambda(\vec{b} \times \vec{c})$.
Calculate $\vec{b} \times \vec{c} = (\hat{j} - \hat{k}) \times (\hat{k} - \hat{i}) = \hat{j} \times \hat{k} - \hat{j} \times \hat{i} - \hat{k} \times \hat{k} + \hat{k} \times \hat{i} = \hat{i} + \hat{j} + \hat{k}$.
So,$\vec{d} = \lambda(\hat{i} + \hat{j} + \hat{k})$.
Since $x=y$,this is consistent. Substituting into the unit vector condition: $\lambda^2(1^2 + 1^2 + 1^2) = 1 \implies 3\lambda^2 = 1 \implies \lambda = \pm \frac{1}{\sqrt{3}}$.
However,checking the condition $\vec{a} \cdot \vec{d} = 0$ with $\vec{d} = \lambda(\hat{i} + \hat{j} + \hat{k})$ gives $(\hat{i} - \hat{j}) \cdot \lambda(\hat{i} + \hat{j} + \hat{k}) = \lambda(1 - 1 + 0) = 0$,which is satisfied for any $\lambda$.
Wait,the vector $\vec{d}$ must be perpendicular to $\vec{a}$ and lie in the plane of $\vec{b}$ and $\vec{c}$. The vector $\vec{b} \times \vec{c} = \hat{i} + \hat{j} + \hat{k}$ is perpendicular to the plane of $\vec{b}$ and $\vec{c}$.
Actually,$\vec{d}$ must be perpendicular to $\vec{a}$ and perpendicular to $(\vec{b} \times \vec{c})$.
Thus,$\vec{d} = \pm \frac{\vec{a} \times (\vec{b} \times \vec{c})}{|\vec{a} \times (\vec{b} \times \vec{c})|}$.
$\vec{a} \times (\vec{b} \times \vec{c}) = (\hat{i} - \hat{j}) \times (\hat{i} + \hat{j} + \hat{k}) = \hat{i} \times \hat{i} + \hat{i} \times \hat{j} + \hat{i} \times \hat{k} - \hat{j} \times \hat{i} - \hat{j} \times \hat{j} - \hat{j} \times \hat{k} = 0 + \hat{k} - \hat{j} + \hat{k} - 0 - \hat{i} = -\hat{i} - \hat{j} + 2\hat{k}$.
The magnitude is $\sqrt{(-1)^2 + (-1)^2 + 2^2} = \sqrt{6}$.
Therefore,$\vec{d} = \pm \frac{-\hat{i} - \hat{j} + 2\hat{k}}{\sqrt{6}} = \pm \frac{\hat{i} + \hat{j} - 2\hat{k}}{\sqrt{6}}$.

(C) Three vectors $\vec{a}, \vec{b}, \vec{c}$ are coplanar if and only if their scalar triple product is zero,i.e.,$[\vec{a} \vec{b} \vec{c}] = 0$.
The scalar triple product is given by the determinant:
$\left|\begin{array}{ccc} -\lambda^2 & 1 & 1 \\ 1 & -\lambda^2 & 1 \\ 1 & 1 & -\lambda^2 \end{array}\right| = 0$
Expanding the determinant along the first row:
$-\lambda^2(\lambda^4 - 1) - 1(-\lambda^2 - 1) + 1(1 + \lambda^2) = 0$
$-\lambda^6 + \lambda^2 + \lambda^2 + 1 + 1 + \lambda^2 = 0$
$-\lambda^6 + 3\lambda^2 + 2 = 0$
$\lambda^6 - 3\lambda^2 - 2 = 0$
Let $x = \lambda^2$. Then the equation becomes $x^3 - 3x - 2 = 0$.
By inspection,$x = -1$ is a root: $(-1)^3 - 3(-1) - 2 = -1 + 3 - 2 = 0$.
Dividing by $(x+1)$,we get $(x+1)(x^2 - x - 2) = 0$,which factors further to $(x+1)(x+1)(x-2) = 0$.
So,$(x+1)^2(x-2) = 0$.
Since $x = \lambda^2$,we have $(\lambda^2 + 1)^2(\lambda^2 - 2) = 0$.
For real $\lambda$,$\lambda^2 + 1$ is always $\ge 1$,so $\lambda^2 + 1 \neq 0$.
Thus,we must have $\lambda^2 - 2 = 0$,which gives $\lambda^2 = 2$.
Therefore,$\lambda = \pm \sqrt{2}$.
There are $2$ distinct real values of $\lambda$.

(C) The volume $V$ of a parallelepiped formed by vectors $\vec{u}, \vec{v}, \vec{w}$ is given by the scalar triple product $|\vec{u} \cdot (\vec{v} \times \vec{w})|$.
Given vectors are $\vec{u} = (1, a, 1)$,$\vec{v} = (0, 1, a)$,and $\vec{w} = (a, 0, 1)$.
$V = |\det(\begin{bmatrix} 1 & a & 1 \\ 0 & 1 & a \\ a & 0 & 1 \end{bmatrix})| = |1(1-0) - a(0-a^2) + 1(0-a)| = |1 + a^3 - a|$.
Let $f(a) = a^3 - a + 1$. To find the minimum volume,we differentiate $f(a)$ with respect to $a$:
$f'(a) = 3a^2 - 1$.
Setting $f'(a) = 0$,we get $3a^2 = 1$,so $a^2 = \frac{1}{3}$,which gives $a = \pm \frac{1}{\sqrt{3}}$.
Since the volume must be non-negative,we consider the magnitude. At $a = \frac{1}{\sqrt{3}}$,$f(a) = (\frac{1}{\sqrt{3}})^3 - \frac{1}{\sqrt{3}} + 1 = \frac{1}{3\sqrt{3}} - \frac{1}{\sqrt{3}} + 1 = 1 - \frac{2}{3\sqrt{3}} > 0$.
Checking the options,the value of $a$ that minimizes the expression is $a = \frac{1}{\sqrt{3}}$.

(D) The scalar triple product is defined as $[\bar{a} \bar{b} \bar{c}] = \bar{a} \cdot (\bar{b} \times \bar{c})$.
Since $\bar{a}, \bar{b}, \bar{c}$ are mutually perpendicular,the vector $(\bar{b} \times \bar{c})$ is parallel to $\bar{a}$.
Let $\bar{n}$ be the unit vector in the direction of $\bar{a}$. Then $\bar{a} = a\bar{n}$.
Since $\bar{b}$ and $\bar{c}$ are perpendicular to each other and to $\bar{a}$,the magnitude of their cross product is $|\bar{b} \times \bar{c}| = |\bar{b}| |\bar{c}| \sin(90^{\circ}) = bc$.
Also,$\bar{b} \times \bar{c}$ points in the direction of $\bar{a}$ (or $-\bar{a}$),so $\bar{b} \times \bar{c} = \pm (bc) \bar{n}$.
Thus,$[\bar{a} \bar{b} \bar{c}] = \bar{a} \cdot (\bar{b} \times \bar{c}) = (a\bar{n}) \cdot (\pm bc \bar{n}) = \pm abc (\bar{n} \cdot \bar{n}) = \pm abc$.
In the context of standard orientation,the magnitude is $abc$.

(A) The volume of a parallelepiped with coterminous edges $\hat{a}, \hat{b}, \hat{c}$ is given by the scalar triple product $|[\hat{a} \hat{b} \hat{c}]|$.
The square of the volume is given by the determinant of the Gram matrix:
$V^2 = \begin{vmatrix} \hat{a} \cdot \hat{a} & \hat{a} \cdot \hat{b} & \hat{a} \cdot \hat{c} \\ \hat{b} \cdot \hat{a} & \hat{b} \cdot \hat{b} & \hat{b} \cdot \hat{c} \\ \hat{c} \cdot \hat{a} & \hat{c} \cdot \hat{b} & \hat{c} \cdot \hat{c} \end{vmatrix}$
Since $\hat{a}, \hat{b}, \hat{c}$ are unit vectors,$\hat{a} \cdot \hat{a} = \hat{b} \cdot \hat{b} = \hat{c} \cdot \hat{c} = 1$. Given $\hat{a} \cdot \hat{b} = \hat{b} \cdot \hat{c} = \hat{c} \cdot \hat{a} = \frac{1}{2}$,we have:
$V^2 = \begin{vmatrix} 1 & 1/2 & 1/2 \\ 1/2 & 1 & 1/2 \\ 1/2 & 1/2 & 1 \end{vmatrix}$
Calculating the determinant:
$V^2 = 1(1 - 1/4) - 1/2(1/2 - 1/4) + 1/2(1/4 - 1/2)$
$V^2 = 1(3/4) - 1/2(1/4) + 1/2(-1/4)$
$V^2 = 3/4 - 1/8 - 1/8 = 3/4 - 2/8 = 3/4 - 1/4 = 2/4 = 1/2$
Therefore,the volume $V = \sqrt{1/2} = \frac{1}{\sqrt{2}}$.

(D) Let the points be $A(1, 2, 2), B(2, 1, 2), C(2, 2, z)$ and $D(1, 1, 1)$.
Four points are coplanar if the volume of the tetrahedron formed by them is $0$. The volume $V$ is given by $\frac{1}{6} |[\vec{AB}, \vec{AC}, \vec{AD}]| = 0$.
Vectors are:
$\vec{AB} = (2-1)\hat{i} + (1-2)\hat{j} + (2-2)\hat{k} = \hat{i} - \hat{j} + 0\hat{k}$
$\vec{AC} = (2-1)\hat{i} + (2-2)\hat{j} + (z-2)\hat{k} = \hat{i} + 0\hat{j} + (z-2)\hat{k}$
$\vec{AD} = (1-1)\hat{i} + (1-2)\hat{j} + (1-2)\hat{k} = 0\hat{i} - \hat{j} - \hat{k}$
The scalar triple product is the determinant:
$\begin{vmatrix} 1 & -1 & 0 \\ 1 & 0 & z-2 \\ 0 & -1 & -1 \end{vmatrix} = 0$
Expanding along the first row:
$1(0 - (-(z-2))) - (-1)(-1 - 0) + 0 = 0$
$1(z-2) - 1 = 0$
$z - 2 - 1 = 0 \implies z = 3$.
Since $z=3$ and not $2$,Statement-$1$ is false. Statement-$2$ is a standard geometric property of coplanar points,so it is true.

(D) Given vectors are $\vec{a} = \hat{i} + \hat{j} + \hat{k}$,$\vec{b} = \hat{i} - \hat{j} + 2\hat{k}$,and $\vec{c} = x\hat{i} + (x-2)\hat{j} - \hat{k}$.
Since $\vec{c}$ lies in the plane of $\vec{a}$ and $\vec{b}$,the scalar triple product $[\vec{a} \vec{b} \vec{c}]$ must be zero.
This implies the determinant of the components is zero:
$\begin{vmatrix} 1 & 1 & 1 \\ 1 & -1 & 2 \\ x & x-2 & -1 \end{vmatrix} = 0$
Expanding the determinant along the first row:
$1((-1)(-1) - 2(x-2)) - 1((1)(-1) - 2(x)) + 1((1)(x-2) - (-1)(x)) = 0$
$1(1 - 2x + 4) - 1(-1 - 2x) + 1(x - 2 + x) = 0$
$(5 - 2x) + (1 + 2x) + (2x - 2) = 0$
$5 - 2x + 1 + 2x + 2x - 2 = 0$
$2x + 4 = 0$
$2x = -4$
$x = -2$

(D) Given the equation: $[3\vec{u}, p\vec{v}, p\vec{w}] - [p\vec{v}, \vec{w}, q\vec{u}] - [2\vec{w}, q\vec{v}, q\vec{u}] = 0$
Using the property of scalar triple product $[k\vec{a}, l\vec{b}, m\vec{c}] = klm[\vec{a}, \vec{b}, \vec{c}]$,we get:
$3p^2[\vec{u}, \vec{v}, \vec{w}] - pq[\vec{v}, \vec{w}, \vec{u}] - 2q^2[\vec{w}, \vec{v}, \vec{u}] = 0$
Since $[\vec{v}, \vec{w}, \vec{u}] = [\vec{u}, \vec{v}, \vec{w}]$ and $[\vec{w}, \vec{v}, \vec{u}] = -[\vec{u}, \vec{v}, \vec{w}]$,the equation becomes:
$3p^2[\vec{u}, \vec{v}, \vec{w}] - pq[\vec{u}, \vec{v}, \vec{w}] + 2q^2[\vec{u}, \vec{v}, \vec{w}] = 0$
$(3p^2 - pq + 2q^2)[\vec{u}, \vec{v}, \vec{w}] = 0$
Since $\vec{u}, \vec{v}, \vec{w}$ are non-coplanar,$[\vec{u}, \vec{v}, \vec{w}] \neq 0$. Therefore:
$3p^2 - pq + 2q^2 = 0$
This is a quadratic equation in $p$. For $p$ to be a real number,the discriminant $D$ must be $\geq 0$:
$D = (-q)^2 - 4(3)(2q^2) = q^2 - 24q^2 = -23q^2$
For $D \geq 0$,we must have $-23q^2 \geq 0$,which implies $q^2 \leq 0$. Since $q$ is a real number,this is only possible if $q = 0$.
Substituting $q = 0$ into $3p^2 - pq + 2q^2 = 0$,we get $3p^2 = 0$,so $p = 0$.
Thus,the only solution is $(p, q) = (0, 0)$.

(B) The scalar triple product is defined as $[\vec{x}, \vec{y}, \vec{z}] = (\vec{x} \times \vec{y}) \cdot \vec{z}$.
Given expression: $[\vec{a} \times \vec{b}, \vec{b} \times \vec{c}, \vec{c} \times \vec{a}]$.
Let $\vec{u} = \vec{a} \times \vec{b}$,$\vec{v} = \vec{b} \times \vec{c}$,and $\vec{w} = \vec{c} \times \vec{a}$.
Then $[\vec{u}, \vec{v}, \vec{w}] = (\vec{u} \times \vec{v}) \cdot \vec{w}$.
Using the vector identity $(\vec{a} \times \vec{b}) \times (\vec{b} \times \vec{c}) = [\vec{a}, \vec{b}, \vec{c}] \vec{b}$,we have:
$(\vec{a} \times \vec{b}) \times (\vec{b} \times \vec{c}) = [\vec{a}, \vec{b}, \vec{c}] \vec{b}$.
Therefore,$[\vec{a} \times \vec{b}, \vec{b} \times \vec{c}, \vec{c} \times \vec{a}] = ((\vec{a} \times \vec{b}) \times (\vec{b} \times \vec{c})) \cdot (\vec{c} \times \vec{a})$.
$= ([\vec{a}, \vec{b}, \vec{c}] \vec{b}) \cdot (\vec{c} \times \vec{a})$.
$= [\vec{a}, \vec{b}, \vec{c}] (\vec{b} \cdot (\vec{c} \times \vec{a}))$.
$= [\vec{a}, \vec{b}, \vec{c}] [\vec{b}, \vec{c}, \vec{a}]$.
Since $[\vec{a}, \vec{b}, \vec{c}] = [\vec{b}, \vec{c}, \vec{a}]$,we get:
$[\vec{a}, \vec{b}, \vec{c}] \cdot [\vec{a}, \vec{b}, \vec{c}] = [\vec{a}, \vec{b}, \vec{c}]^2$.
Comparing this with $\lambda [\vec{a}, \vec{b}, \vec{c}]^2$,we find $\lambda = 1$.

(D) Given that the vectors $a, b, c$ are linearly dependent,their scalar triple product must be zero: $[a, b, c] = 0$.
This implies the determinant of the matrix formed by their components is zero:
$\begin{vmatrix} 1 & 1 & 1 \\ 4 & 3 & 4 \\ 1 & \alpha & \beta \end{vmatrix} = 0$
Expanding the determinant along the first row:
$1(3\beta - 4\alpha) - 1(4\beta - 4) + 1(4\alpha - 3) = 0$
$3\beta - 4\alpha - 4\beta + 4 + 4\alpha - 3 = 0$
$-\beta + 1 = 0 \Rightarrow \beta = 1$.
Given $|c| = \sqrt{3}$,we have $|c|^2 = 3$.
$1^2 + \alpha^2 + \beta^2 = 3$
$1 + \alpha^2 + 1^2 = 3$
$\alpha^2 + 2 = 3 \Rightarrow \alpha^2 = 1 \Rightarrow \alpha = \pm 1$.
Thus,$\alpha = \pm 1$ and $\beta = 1$.

(B) Let $V = a \cdot (b \times c) = [a, b, c]$.
Given $p = \frac{b \times c}{V}$,$q = \frac{c \times a}{V}$,$r = \frac{a \times b}{V}$.
Then $[p, q, r] = p \cdot (q \times r) = \frac{b \times c}{V} \cdot \left( \frac{c \times a}{V} \times \frac{a \times b}{V} \right)$.
Using the vector identity $(c \times a) \times (a \times b) = [c, a, b] a - [c, a, a] b = [a, b, c] a - 0 = V a$.
Thus,$[p, q, r] = \frac{1}{V^3} (b \times c) \cdot (V a) = \frac{V}{V^3} (b \times c) \cdot a = \frac{V^2}{V^3} = \frac{1}{V} = \frac{1}{a \cdot (b \times c)}$.

(A) We need to evaluate the expression $(a + b + c) \cdot [(a + b) \times (a + c)]$.
First,expand the cross product: $(a + b) \times (a + c) = a \times a + a \times c + b \times a + b \times c$.
Since $a \times a = 0$,this simplifies to $a \times c + b \times a + b \times c$.
Now,take the dot product with $(a + b + c)$:
$(a + b + c) \cdot (a \times c + b \times a + b \times c) = [a, a, c] + [a, b, a] + [a, b, c] + [b, a, c] + [b, b, a] + [b, b, c] + [c, a, c] + [c, b, a] + [c, b, c]$.
Using the property that the scalar triple product is zero if any two vectors are identical,we have:
$[a, a, c] = 0, [a, b, a] = 0, [b, b, a] = 0, [b, b, c] = 0, [c, a, c] = 0, [c, b, c] = 0$.
This leaves us with $[a, b, c] + [b, a, c] + [c, b, a]$.
Since $[b, a, c] = -[a, b, c]$ and $[c, b, a] = [a, b, c]$,the expression becomes:
$[a, b, c] - [a, b, c] + [a, b, c] = [a, b, c]$.
Wait,let us re-evaluate: $[a, b, c] + [b, a, c] + [c, b, a] = [a, b, c] - [a, b, c] + [a, b, c] = [a, b, c]$.
Correction: The expansion is $(a+b+c) \cdot (a \times c + b \times a + b \times c) = [a, a, c] + [a, b, a] + [a, b, c] + [b, a, c] + [b, b, a] + [b, b, c] + [c, a, c] + [c, b, a] + [c, b, c] = 0 + 0 + [a, b, c] - [a, b, c] + 0 + 0 + 0 + [a, b, c] + 0 = [a, b, c]$.

(C) The scalar triple product is defined as $[U V W] = U \cdot (V \times W)$.
First,we calculate the cross product $V \times W$:
$V \times W = \begin{vmatrix} i & j & k \\ 2 & 1 & -1 \\ 1 & 0 & 3 \end{vmatrix} = i(3 - 0) - j(6 - (-1)) + k(0 - 1) = 3i - 7j - k$.
Let $A = V \times W = 3i - 7j - k$.
The scalar triple product is $[U V W] = U \cdot A = |U| |A| \cos \theta$,where $\theta$ is the angle between $U$ and $A$.
Since $U$ is a unit vector,$|U| = 1$.
Thus,$[U V W] = |A| \cos \theta$.
The maximum value occurs when $\cos \theta = 1$,which is $|A|$.
$|A| = \sqrt{3^2 + (-7)^2 + (-1)^2} = \sqrt{9 + 49 + 1} = \sqrt{59}$.
Therefore,the maximum value is $\sqrt{59}$.

(D) For three vectors to be coplanar,their scalar triple product must be zero.
$\begin{vmatrix} 1-x & 1 & 1 \\ 1 & 1-y & 1 \\ 1 & 1 & 1-z \end{vmatrix} = 0$
Subtracting the first row from the second and third rows:
$\begin{vmatrix} 1-x & 1 & 1 \\ x & -y & 0 \\ x & 0 & -z \end{vmatrix} = 0$
Expanding along the first row:
$(1-x)(yz) - 1(-xz) + 1(xz) = 0$
$yz - xyz + xz + xz = 0$
$yz - xyz + 2xz = 0$
Dividing by $xyz$ (since $x, y, z \neq 0$):
$\frac{1}{x} - 1 + \frac{2}{y} = 0$ (Wait,let us re-evaluate the determinant expansion).
Correct expansion: $(1-x)((1-y)(1-z) - 1) - 1(1-z-1) + 1(1-(1-y)) = 0$
$(1-x)(1-z-y+yz-1) + z + y = 0$
$(1-x)(yz-y-z) + y + z = 0$
$yz - y - z - xyz + xy + xz + y + z = 0$
$yz - xyz + xy + xz = 0$
Dividing by $xyz$: $\frac{1}{x} + \frac{1}{z} + \frac{1}{y} = 1$.

(C) The given determinant is the Gramian determinant of the vectors $\vec{a}, \vec{b}, \vec{c}$,which is equal to the square of the scalar triple product $[\vec{a} \vec{b} \vec{c}]^2$.
First,we calculate the scalar triple product $[\vec{a} \vec{b} \vec{c}] = \vec{a} \cdot (\vec{b} \times \vec{c}) = \left| \begin{matrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 2 & -1 \end{matrix} \right|$.
Expanding the determinant along the first row:
$[\vec{a} \vec{b} \vec{c}] = 1((-1)(-1) - (1)(2)) - 1((1)(-1) - (1)(1)) + 1((1)(2) - (-1)(1))$
$= 1(1 - 2) - 1(-1 - 1) + 1(2 + 1)$
$= 1(-1) - 1(-2) + 1(3)$
$= -1 + 2 + 3 = 4$.
Therefore,the value of the determinant is $[\vec{a} \vec{b} \vec{c}]^2 = 4^2 = 16$.

(D) Given the scalar triple product: $(\vec{a} - \lambda \vec{b}) \cdot ((\vec{b} - 2\vec{c}) \times (\vec{c} + 2\vec{a})) = 0$.
First,expand the cross product: $(\vec{b} - 2\vec{c}) \times (\vec{c} + 2\vec{a}) = \vec{b} \times \vec{c} + 2(\vec{b} \times \vec{a}) - 2(\vec{c} \times \vec{c}) - 4(\vec{c} \times \vec{a})$.
Since $\vec{c} \times \vec{c} = 0$,this simplifies to $\vec{b} \times \vec{c} + 2(\vec{b} \times \vec{a}) - 4(\vec{c} \times \vec{a})$.
Now,take the dot product with $(\vec{a} - \lambda \vec{b})$:
$(\vec{a} - \lambda \vec{b}) \cdot (\vec{b} \times \vec{c} + 2(\vec{b} \times \vec{a}) - 4(\vec{c} \times \vec{a})) = 0$.
Distributing the dot product:
$\vec{a} \cdot (\vec{b} \times \vec{c}) + 2\vec{a} \cdot (\vec{b} \times \vec{a}) - 4\vec{a} \cdot (\vec{c} \times \vec{a}) - \lambda \vec{b} \cdot (\vec{b} \times \vec{c}) - 2\lambda \vec{b} \cdot (\vec{b} \times \vec{a}) + 4\lambda \vec{b} \cdot (\vec{c} \times \vec{a}) = 0$.
Using the property that the scalar triple product is zero if any two vectors are the same,we have $\vec{a} \cdot (\vec{b} \times \vec{a}) = 0$,$\vec{a} \cdot (\vec{c} \times \vec{a}) = 0$,$\vec{b} \cdot (\vec{b} \times \vec{c}) = 0$,and $\vec{b} \cdot (\vec{b} \times \vec{a}) = 0$.
This leaves: $\vec{a} \cdot (\vec{b} \times \vec{c}) + 4\lambda \vec{b} \cdot (\vec{c} \times \vec{a}) = 0$.
Since $\vec{b} \cdot (\vec{c} \times \vec{a}) = \vec{a} \cdot (\vec{b} \times \vec{c})$,we get: $(1 + 4\lambda) [\vec{a} \vec{b} \vec{c}] = 0$.
Since $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar,$[\vec{a} \vec{b} \vec{c}] \neq 0$,so $1 + 4\lambda = 0$,which implies $\lambda = -1/4$.

(D) Since $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ are non-coplanar,the vectors $\overrightarrow{P}, \overrightarrow{Q}, \overrightarrow{R}$ are linearly dependent if and only if their scalar triple product is zero,i.e.,$[\overrightarrow{P}, \overrightarrow{Q}, \overrightarrow{R}] = 0$.
This is equivalent to the determinant of the coefficients being zero:
$\begin{vmatrix} 1 & 1 & 1 \\ 4 & 3 & 4 \\ 1 & \alpha & \beta \end{vmatrix} = 0$
Expanding the determinant along the first row:
$1(3\beta - 4\alpha) - 1(4\beta - 4) + 1(4\alpha - 3) = 0$
$3\beta - 4\alpha - 4\beta + 4 + 4\alpha - 3 = 0$
$-\beta + 1 = 0$
$\beta = 1$
Since the equation simplifies to $\beta = 1$ and is independent of $\alpha$,$\alpha$ can take any real value. Therefore,there are infinitely many possible values for $\alpha$.

(A) The volume of a parallelepiped with coterminous edges $\vec{a}, \vec{b}, \vec{c}$ is given by the scalar triple product $[\vec{a} \vec{b} \vec{c}] = 12$.
The volume of a tetrahedron with coterminous edges $\vec{u}, \vec{v}, \vec{w}$ is given by $V = \frac{1}{6} |[\vec{u} \vec{v} \vec{w}]|$.
Here,$\vec{u} = \vec{a} - \vec{b}$,$\vec{v} = \vec{b} - \vec{c}$,and $\vec{w} = \vec{a} + \vec{b} - \vec{c}$.
Calculating the scalar triple product $[\vec{u} \vec{v} \vec{w}] = [(\vec{a} - \vec{b}) (\vec{b} - \vec{c}) (\vec{a} + \vec{b} - \vec{c})]$.
Using the properties of the scalar triple product,this simplifies to:
$[\vec{a} - \vec{b}, \vec{b} - \vec{c}, \vec{a} + \vec{b} - \vec{c}] = [\vec{a} \vec{b} \vec{c}] = 12$.
Therefore,the volume of the tetrahedron is $\frac{1}{6} \times 12 = 2$ cubic units.

(B) Let the vertices of the tetrahedron be $A(\vec{0}), B(\vec{a}), C(\vec{b}),$ and $D(\vec{c})$.
The volume of tetrahedron $ABCD$ is given by $V = \frac{1}{6} |[\vec{a} \vec{b} \vec{c}]| = 81$.
The centroids of the faces are:
$G_1 = \frac{\vec{a} + \vec{b}}{3}$
$G_2 = \frac{\vec{a} + \vec{c}}{3}$
$G_3 = \frac{\vec{b} + \vec{c}}{3}$
The volume of tetrahedron $AG_1G_2G_3$ is given by $V' = \frac{1}{6} |[\vec{G_1} \vec{G_2} \vec{G_3}]|$.
Substituting the values:
$V' = \frac{1}{6} |[\frac{\vec{a} + \vec{b}}{3}, \frac{\vec{a} + \vec{c}}{3}, \frac{\vec{b} + \vec{c}}{3}]|$
Using the properties of the scalar triple product:
$V' = \frac{1}{6} \times \frac{1}{27} |[\vec{a} + \vec{b}, \vec{a} + \vec{c}, \vec{b} + \vec{c}]|$
Expanding the scalar triple product:
$[\vec{a} + \vec{b}, \vec{a} + \vec{c}, \vec{b} + \vec{c}] = 2[\vec{a} \vec{b} \vec{c}]$
Therefore,$V' = \frac{1}{6} \times \frac{1}{27} \times 2 |[\vec{a} \vec{b} \vec{c}]| = \frac{2}{27} \times 81 = 6$ cubic units.

(A) Since the vectors $\vec a, \vec b$,and $\vec c$ are coplanar,their scalar triple product must be zero: $[\vec a \vec b \vec c] = 0$.
This is given by the determinant:
$\begin{vmatrix} 2\sin \theta & -1 & 2 \\ 2 & 2\sin \theta & -1 \\ 4 & -1 & 4\cos^2 \theta \end{vmatrix} = 0$
Expanding the determinant along the first row:
$2\sin \theta (8\sin \theta \cos^2 \theta - 1) + 1(8\cos^2 \theta + 4) + 2(-2 - 8\sin \theta) = 0$
$16\sin^2 \theta \cos^2 \theta - 2\sin \theta + 8\cos^2 \theta + 4 - 4 - 16\sin \theta = 0$
$4\sin^2(2\theta) + 8(1 - \sin^2 \theta) - 18\sin \theta = 0$
$4(4\sin^2 \theta \cos^2 \theta) + 8 - 8\sin^2 \theta - 18\sin \theta = 0$
Solving this equation,we find that $\sin \theta = \frac{1}{2}$ satisfies the condition.
For $\sin \theta = \frac{1}{2}$,the general solution is $\theta = n\pi + {(-1)^n}\frac{\pi}{6}$ for $n \in I$.