Scalar triple product and their applications Questions in English

(B) For four points $A, B, C, D$ to be coplanar,the scalar triple product of vectors $\vec{AB}, \vec{AC},$ and $\vec{AD}$ must be zero,i.e.,$[\vec{AB} \ \vec{AC} \ \vec{AD}] = 0$.
First,we find the vectors:
$\vec{AB} = (0-2)\hat{i} + (-1-1)\hat{j} + (0-(-1))\hat{k} = -2\hat{i} - 2\hat{j} + \hat{k}$
$\vec{AC} = (4-2)\hat{i} + (0-1)\hat{j} + (4-(-1))\hat{k} = 2\hat{i} - \hat{j} + 5\hat{k}$
$\vec{AD} = (2-2)\hat{i} + (0-1)\hat{j} + (x-(-1))\hat{k} = 0\hat{i} - \hat{j} + (x+1)\hat{k}$
The condition for coplanarity is given by the determinant:
$\begin{vmatrix} -2 & -2 & 1 \\ 2 & -1 & 5 \\ 0 & -1 & x+1 \end{vmatrix} = 0$
Expanding the determinant along the first row:
$-2[(-1)(x+1) - (5)(-1)] - (-2)[(2)(x+1) - (5)(0)] + 1[(2)(-1) - (-1)(0)] = 0$
$-2[-x-1+5] + 2[2x+2] + 1[-2] = 0$
$-2[4-x] + 4x + 4 - 2 = 0$
$-8 + 2x + 4x + 2 = 0$
$6x - 6 = 0$
$6x = 6$
$x = 1$

(D) Since the three vectors are coplanar,their scalar triple product must be zero.
The determinant formed by the components of these vectors is:
$\begin{vmatrix} a & a & c \\ 1 & 0 & 1 \\ c & c & b \end{vmatrix} = 0$
Applying the column operation $C_1 \rightarrow C_1 - C_2$:
$\begin{vmatrix} 0 & a & c \\ 1 & 0 & 1 \\ 0 & c & b \end{vmatrix} = 0$
Expanding along the first column $(C_1)$:
$-1(ab - c^2) = 0$
$c^2 - ab = 0 \Rightarrow c^2 = ab$
This implies that $c$ is the Geometric Mean ($G$.$M$.) of $a$ and $b$.

(A) To determine if the vectors are coplanar,we calculate the scalar triple product $[\bar{a} \ \bar{b} \ \bar{c}]$.
If the scalar triple product is $0$,the vectors are coplanar. If it is non-zero,they are non-coplanar.
Given $\bar{a} = 3\hat{\imath} + \hat{\jmath} - \hat{k}$,$\bar{b} = 2\hat{\imath} - \hat{\jmath} + 7\hat{k}$,and $\bar{c} = 7\hat{\imath} - \hat{\jmath} + 23\hat{k}$.
$[\bar{a} \ \bar{b} \ \bar{c}] = \begin{vmatrix} 3 & 1 & -1 \\ 2 & -1 & 7 \\ 7 & -1 & 23 \end{vmatrix}$
$= 3((-1)(23) - (7)(-1)) - 1((2)(23) - (7)(7)) - 1((2)(-1) - (7)(-1))$
$= 3(-23 + 7) - 1(46 - 49) - 1(-2 + 7)$
$= 3(-16) - 1(-3) - 1(5)$
$= -48 + 3 - 5 = -50$.
Since the scalar triple product is $-50 \neq 0$,the vectors $\bar{a}, \bar{b},$ and $\bar{c}$ are non-coplanar. Thus,statement $A$ is true.

(C) If the points $O(0,0,0)$,$P(2,3,4)$,$Q(1,2,3)$,and $R(x, y, z)$ are co-planar,then the scalar triple product of the vectors $\vec{OR}$,$\vec{OP}$,and $\vec{OQ}$ must be zero.
$\implies [\vec{OR} \quad \vec{OP} \quad \vec{OQ}] = 0$
Using the determinant form:
$\left| \begin{array}{ccc} x & y & z \\ 2 & 3 & 4 \\ 1 & 2 & 3 \end{array} \right| = 0$
Expanding along the first row:
$x(3 \times 3 - 4 \times 2) - y(2 \times 3 - 4 \times 1) + z(2 \times 2 - 3 \times 1) = 0$
$x(9 - 8) - y(6 - 4) + z(4 - 3) = 0$
$x(1) - y(2) + z(1) = 0$
$x - 2y + z = 0$
Thus,the correct equation is $x - 2y + z = 0$.

(D) Let $\vec{a} = 2\hat{i} - \hat{j} - \hat{k}$,$\vec{b} = \hat{i} + 2\hat{j} - 3\hat{k}$ and $\vec{c} = 3\hat{i} + \lambda\hat{j} + 5\hat{k}$.
Since these vectors are coplanar,their scalar triple product must be zero,i.e.,$[\vec{a} \vec{b} \vec{c}] = 0$.
This implies the determinant of the components is zero:
$\begin{vmatrix} 2 & -1 & -1 \\ 1 & 2 & -3 \\ 3 & \lambda & 5 \end{vmatrix} = 0$
Expanding the determinant along the first row:
$2(2(5) - (-3)(\lambda)) - (-1)(1(5) - (-3)(3)) + (-1)(1(\lambda) - 2(3)) = 0$
$2(10 + 3\lambda) + 1(5 + 9) - 1(\lambda - 6) = 0$
$20 + 6\lambda + 14 - \lambda + 6 = 0$
$5\lambda + 40 = 0$
$5\lambda = -40$
$\lambda = -8$.

(B) Since the vectors $a$,$b$,and $c$ are coplanar,there must exist scalars $x$,$y$,and $z$ (not all zero) such that $x a + y b + z c = 0$ $(i)$.
Multiplying both sides of equation $(i)$ by $a$ and $b$ respectively,we get:
$x(a \cdot a) + y(a \cdot b) + z(a \cdot c) = 0$ $(ii)$
$x(b \cdot a) + y(b \cdot b) + z(b \cdot c) = 0$ $(iii)$
Considering equations $(i)$,$(ii)$,and $(iii)$ as a system of linear equations in variables $x$,$y$,and $z$,since a non-trivial solution exists,the determinant of the coefficient matrix must be zero.
Therefore,$\left|\begin{array}{ccc} a & b & c \\ a \cdot a & a \cdot b & a \cdot c \\ b \cdot a & b \cdot b & b \cdot c \end{array}\right| = 0$.

(D) Since the three given vectors are coplanar,their scalar triple product must be zero. Alternatively,one vector can be expressed as a linear combination of the other two. Let $\overrightarrow{u} = \overrightarrow{a}+\lambda \overrightarrow{b}+3 \overrightarrow{c}$,$\overrightarrow{v} = -2 \overrightarrow{a}+3 \overrightarrow{b}-4 \overrightarrow{c}$,and $\overrightarrow{w} = \overrightarrow{a}-3 \overrightarrow{b}+5 \overrightarrow{c}$.
For coplanar vectors,the determinant of their coefficients must be zero:
$\begin{vmatrix} 1 & \lambda & 3 \\ -2 & 3 & -4 \\ 1 & -3 & 5 \end{vmatrix} = 0$
Expanding 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) Let $\bar{v} = (2 \bar{a} - \bar{b}) \cdot [(\bar{a} \times \bar{b}) \times (\bar{a} + 2 \bar{b})]$.
Using the vector triple product identity $(\bar{x} \times \bar{y}) \times \bar{z} = (\bar{x} \cdot \bar{z}) \bar{y} - (\bar{y} \cdot \bar{z}) \bar{x}$,we have:
$(\bar{a} \times \bar{b}) \times (\bar{a} + 2 \bar{b}) = (\bar{a} \cdot (\bar{a} + 2 \bar{b})) \bar{b} - (\bar{b} \cdot (\bar{a} + 2 \bar{b})) \bar{a}$.
Since $\bar{a}$ and $\bar{b}$ are unit vectors $(|\bar{a}| = 1, |\bar{b}| = 1)$,let $\bar{a} \cdot \bar{b} = \cos \theta$.
Then $\bar{a} \cdot \bar{a} = 1$ and $\bar{b} \cdot \bar{b} = 1$.
So,$(\bar{a} \times \bar{b}) \times (\bar{a} + 2 \bar{b}) = (1 + 2 \cos \theta) \bar{b} - (\cos \theta + 2) \bar{a}$.
Now,$(2 \bar{a} - \bar{b}) \cdot [(1 + 2 \cos \theta) \bar{b} - (\cos \theta + 2) \bar{a}] = 2(1 + 2 \cos \theta)(\bar{a} \cdot \bar{b}) - 2(\cos \theta + 2)(\bar{a} \cdot \bar{a}) - (1 + 2 \cos \theta)(\bar{b} \cdot \bar{b}) + (\cos \theta + 2)(\bar{b} \cdot \bar{a})$.
$= 2(1 + 2 \cos \theta) \cos \theta - 2(\cos \theta + 2) - (1 + 2 \cos \theta) + (\cos \theta + 2) \cos \theta$.
$= 2 \cos \theta + 4 \cos^2 \theta - 2 \cos \theta - 4 - 1 - 2 \cos \theta + \cos^2 \theta + 2 \cos \theta$.
$= 5 \cos^2 \theta - 5 = 5(\cos^2 \theta - 1) = -5 \sin^2 \theta$.
Given $\bar{a} = \frac{1}{\sqrt{10}}(3 \hat{i} + \hat{k})$ and $\bar{b} = \frac{1}{7}(2 \hat{i} + 3 \hat{j} - 6 \hat{k})$,$\cos \theta = \bar{a} \cdot \bar{b} = \frac{1}{7\sqrt{10}}(6 + 0 - 6) = 0$.
Thus,$\sin^2 \theta = 1 - 0^2 = 1$.
The value is $-5(1) = -5$.

(D) Given $\bar{a} = \hat{i} - \hat{j}$,$\bar{b} = \hat{j} - \hat{k}$,and $\bar{c} = \hat{k} - \hat{i}$.
Since $[\bar{b} \bar{c} \bar{d}] = 0$,the vector $\bar{d}$ must be coplanar with $\bar{b}$ and $\bar{c}$.
Thus,$\bar{d} = x\bar{b} + y\bar{c} = x(\hat{j} - \hat{k}) + y(\hat{k} - \hat{i}) = -y\hat{i} + x\hat{j} + (y - x)\hat{k}$.
Given $\bar{a} \cdot \bar{d} = 0$,we have $(\hat{i} - \hat{j}) \cdot (-y\hat{i} + x\hat{j} + (y - x)\hat{k}) = 0$.
$-y - x = 0 \implies y = -x$.
Substituting $y = -x$ into $\bar{d}$,we get $\bar{d} = x\hat{i} + x\hat{j} - 2x\hat{k} = x(\hat{i} + \hat{j} - 2\hat{k})$.
Since $\bar{d}$ is a unit vector,$|\bar{d}| = 1 \implies |x| \sqrt{1^2 + 1^2 + (-2)^2} = 1 \implies |x| \sqrt{6} = 1 \implies x = \pm \frac{1}{\sqrt{6}}$.
Therefore,$\bar{d} = \pm \frac{1}{\sqrt{6}}(\hat{i} + \hat{j} - 2\hat{k})$.

(B) We need to evaluate the scalar triple product: $(\overline{a}+\overline{b}+\overline{c}) \cdot [(\overline{a}+\overline{b}) \times (\overline{a}+\overline{c})]$.
First,expand the cross product: $(\overline{a}+\overline{b}) \times (\overline{a}+\overline{c}) = \overline{a} \times \overline{a} + \overline{a} \times \overline{c} + \overline{b} \times \overline{a} + \overline{b} \times \overline{c} = 0 + \overline{a} \times \overline{c} + \overline{b} \times \overline{a} + \overline{b} \times \overline{c}$.
Now,take the dot product with $(\overline{a}+\overline{b}+\overline{c})$:
$(\overline{a}+\overline{b}+\overline{c}) \cdot (\overline{a} \times \overline{c} + \overline{b} \times \overline{a} + \overline{b} \times \overline{c})$.
Using the property of scalar triple products $[\overline{x} \overline{y} \overline{z}] = \overline{x} \cdot (\overline{y} \times \overline{z})$:
$= [\overline{a} \overline{a} \overline{c}] + [\overline{a} \overline{b} \overline{a}] + [\overline{a} \overline{b} \overline{c}] + [\overline{b} \overline{a} \overline{c}] + [\overline{b} \overline{b} \overline{a}] + [\overline{b} \overline{b} \overline{c}] + [\overline{c} \overline{a} \overline{c}] + [\overline{c} \overline{b} \overline{a}] + [\overline{c} \overline{b} \overline{c}]$.
Since any scalar triple product with two identical vectors is $0$:
$= 0 + 0 + [\overline{a} \overline{b} \overline{c}] + [\overline{b} \overline{a} \overline{c}] + 0 + 0 + 0 + [\overline{c} \overline{b} \overline{a}] + 0$.
Using the property $[\overline{x} \overline{y} \overline{z}] = -[\overline{y} \overline{x} \overline{z}]$:
$= [\overline{a} \overline{b} \overline{c}] - [\overline{a} \overline{b} \overline{c}] + [\overline{a} \overline{b} \overline{c}] = [\overline{a} \overline{b} \overline{c}]$.
Wait,re-evaluating the expansion: $(\overline{a}+\overline{b}+\overline{c}) \cdot (\overline{a} \times \overline{c} + \overline{b} \times \overline{a} + \overline{b} \times \overline{c}) = [\overline{a} \overline{a} \overline{c}] + [\overline{a} \overline{b} \overline{a}] + [\overline{a} \overline{b} \overline{c}] + [\overline{b} \overline{a} \overline{c}] + [\overline{b} \overline{b} \overline{a}] + [\overline{b} \overline{b} \overline{c}] + [\overline{c} \overline{a} \overline{c}] + [\overline{c} \overline{b} \overline{a}] + [\overline{c} \overline{b} \overline{c}] = 0 + 0 + [\overline{a} \overline{b} \overline{c}] - [\overline{a} \overline{b} \overline{c}] + 0 + 0 + 0 + [\overline{a} \overline{b} \overline{c}] + 0 = [\overline{a} \overline{b} \overline{c}]$.
Actually,the correct result is $[\overline{a} \overline{b} \overline{c}]$. Let's re-check the expansion: $(\overline{a}+\overline{b}+\overline{c}) \cdot (\overline{a} \times \overline{c} + \overline{b} \times \overline{a} + \overline{b} \times \overline{c}) = [\overline{a} \overline{a} \overline{c}] + [\overline{a} \overline{b} \overline{a}] + [\overline{a} \overline{b} \overline{c}] + [\overline{b} \overline{a} \overline{c}] + [\overline{b} \overline{b} \overline{a}] + [\overline{b} \overline{b} \overline{c}] + [\overline{c} \overline{a} \overline{c}] + [\overline{c} \overline{b} \overline{a}] + [\overline{c} \overline{b} \overline{c}] = 0 + 0 + [\overline{a} \overline{b} \overline{c}] - [\overline{a} \overline{b} \overline{c}] + 0 + 0 + 0 + [\overline{a} \overline{b} \overline{c}] + 0 = [\overline{a} \overline{b} \overline{c}]$.
Correction: The original provided solution was $-[\overline{a} \overline{b} \overline{c}]$,but the calculation yields $[\overline{a} \overline{b} \overline{c}]$. Re-calculating: $[\overline{a} \overline{b} \overline{c}] + [\overline{b} \overline{a} \overline{c}] + [\overline{c} \overline{b} \overline{a}] = [\overline{a} \overline{b} \overline{c}] - [\overline{a} \overline{b} \overline{c}] + [\overline{a} \overline{b} \overline{c}] = [\overline{a} \overline{b} \overline{c}]$.

(C) Given $\hat{a}=\frac{1}{\sqrt{10}}(3 \hat{i}+\hat{k})$ and $\hat{b}=\frac{1}{7}(2 \hat{i}+3 \hat{j}-6 \hat{k})$.
First,calculate the dot product $\hat{a} \cdot \hat{b} = \frac{1}{7\sqrt{10}}(3 \times 2 + 0 \times 3 + 1 \times (-6)) = \frac{1}{7\sqrt{10}}(6 - 6) = 0$.
Since $\hat{a} \cdot \hat{b} = 0$,$\hat{a}$ and $\hat{b}$ are perpendicular unit vectors,so $|\hat{a}| = 1, |\hat{b}| = 1$ and $\hat{a} \times \hat{b}$ is a unit vector perpendicular to both.
We need to evaluate $(2 \hat{a}-\hat{b}) \cdot [(\hat{a} \times \hat{b}) \times (\hat{a}+2 \hat{b})]$.
Using the scalar triple product property $[\vec{u} \quad \vec{v} \quad \vec{w}] = \vec{u} \cdot (\vec{v} \times \vec{w})$,the expression is $[2 \hat{a}-\hat{b} \quad \hat{a} \times \hat{b} \quad \hat{a}+2 \hat{b}]$.
Using the property of scalar triple product $[\vec{u} \quad \vec{v} \quad \vec{w}] = -[\vec{v} \quad \vec{u} \quad \vec{w}]$,we get $-[\hat{a} \times \hat{b} \quad 2 \hat{a}-\hat{b} \quad \hat{a}+2 \hat{b}]$.
Expanding the cross product inside: $(2 \hat{a}-\hat{b}) \times (\hat{a}+2 \hat{b}) = 2(\hat{a} \times \hat{a}) + 4(\hat{a} \times \hat{b}) - (\hat{b} \times \hat{a}) - 2(\hat{b} \times \hat{b})$.
Since $\hat{a} \times \hat{a} = 0$,$\hat{b} \times \hat{b} = 0$,and $\hat{b} \times \hat{a} = -(\hat{a} \times \hat{b})$,this becomes $0 + 4(\hat{a} \times \hat{b}) + (\hat{a} \times \hat{b}) - 0 = 5(\hat{a} \times \hat{b})$.
Thus,the expression is $-(\hat{a} \times \hat{b}) \cdot [5(\hat{a} \times \hat{b})] = -5 |\hat{a} \times \hat{b}|^2$.
Since $\hat{a} \perp \hat{b}$ and they are unit vectors,$|\hat{a} \times \hat{b}| = |\hat{a}| |\hat{b}| \sin(90^{\circ}) = 1 \times 1 \times 1 = 1$.
Therefore,$-5(1)^2 = -5$.

(D) Let $\bar{r} = a\hat{i} + b\hat{j} + c\hat{k}$.
Since $\bar{r}$ is perpendicular to $\bar{q}$,we have $\bar{r} \cdot \bar{q} = 0$.
This implies $a - 2b + c = 0$ ... $(i)$.
Since $\bar{r}$ is coplanar with $\bar{p}$ and $\bar{q}$,the scalar triple product $[\bar{p} \ \bar{q} \ \bar{r}] = 0$.
$\begin{vmatrix} 1 & 1 & 1 \\ 1 & -2 & 1 \\ a & b & c \end{vmatrix} = 0$.
Expanding the determinant: $1(-2c - b) - 1(c - a) + 1(b + 2a) = 0$.
$-2c - b - c + a + b + 2a = 0 \Rightarrow 3a - 3c = 0 \Rightarrow a = c$ ... $(ii)$.
Substituting $(ii)$ into $(i)$: $a - 2b + a = 0 \Rightarrow 2a = 2b \Rightarrow a = b$.
Thus,$\bar{r} = a(\hat{i} + \hat{j} + \hat{k})$.
The unit vector in the direction of $\bar{r}$ is $\frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}$.
The required vector has magnitude $5\sqrt{3}$,so $\text{Vector} = 5\sqrt{3} \times \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}} = 5(\hat{i} + \hat{j} + \hat{k})$.

(C) Given $|(\vec{a} \times \vec{b}) \cdot \vec{c}| = |\vec{a}||\vec{b}||\vec{c}|$.
Since $|(\vec{a} \times \vec{b}) \cdot \vec{c}| = |\vec{a} \times \vec{b}| |\vec{c}| |\cos \theta|$,where $\theta$ is the angle between $(\vec{a} \times \vec{b})$ and $\vec{c}$,we have $|\vec{a} \times \vec{b}| |\vec{c}| |\cos \theta| = |\vec{a}||\vec{b}||\vec{c}|$.
Substituting $|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}| \sin \phi$,where $\phi$ is the angle between $\vec{a}$ and $\vec{b}$,we get $|\vec{a}||\vec{b}| \sin \phi |\vec{c}| |\cos \theta| = |\vec{a}||\vec{b}||\vec{c}|$.
This implies $\sin \phi = 1$ and $|\cos \theta| = 1$.
Thus,$\phi = 90^{\circ}$ (meaning $\vec{a} \perp \vec{b}$) and $\theta = 0^{\circ}$ or $180^{\circ}$ (meaning $\vec{c}$ is parallel to the normal of the plane containing $\vec{a}$ and $\vec{b}$).
Since $\vec{c}$ is parallel to $\vec{a} \times \vec{b}$,it follows that $\vec{c} \perp \vec{a}$ and $\vec{c} \perp \vec{b}$.
Therefore,$\vec{a} \cdot \vec{b} = 0$,$\vec{b} \cdot \vec{c} = 0$,and $\vec{c} \cdot \vec{a} = 0$.

(A) First,calculate the cross product $\bar{b} \times \bar{d}$:
$\bar{b} \times \bar{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & 3 \\ 1 & -1 & 0 \end{vmatrix} = \hat{i}(0 - (-3)) - \hat{j}(0 - 3) + \hat{k}(-2 - 0) = 3\hat{i} + 3\hat{j} - 2\hat{k}$
Next,calculate the vector $\bar{c} - \bar{a}$:
$\bar{c} - \bar{a} = (4\hat{i} - \hat{j} + 2\hat{k}) - (\hat{i} + 5\hat{k}) = 3\hat{i} - \hat{j} - 3\hat{k}$
Finally,calculate the dot product $(\bar{c} - \bar{a}) \cdot (\bar{b} \times \bar{d})$:
$(\bar{c} - \bar{a}) \cdot (\bar{b} \times \bar{d}) = (3\hat{i} - \hat{j} - 3\hat{k}) \cdot (3\hat{i} + 3\hat{j} - 2\hat{k})$
$= (3)(3) + (-1)(3) + (-3)(-2) = 9 - 3 + 6 = 12$

(A) The volume of a tetrahedron with co-terminus edges $\bar{a}, \bar{b}, \bar{c}$ is given by $V = \frac{1}{6} |[\bar{a} \bar{b} \bar{c}]|$.
Given $V = 12$,we have $\frac{1}{6} |[\bar{a} \bar{b} \bar{c}]| = 12$,which implies $|[\bar{a} \bar{b} \bar{c}]| = 72$.
The scalar triple product is defined as $[\bar{a} \bar{b} \bar{c}] = \bar{a} \cdot (\bar{b} \times \bar{c})$.
The scalar projection of $\bar{a}$ on $\bar{b} \times \bar{c}$ is given by $\frac{\bar{a} \cdot (\bar{b} \times \bar{c})}{|\bar{b} \times \bar{c}|} = 4$.
Let $X = |\bar{b} \times \bar{c}|$. Then $\frac{[\bar{a} \bar{b} \bar{c}]}{X} = 4$,so $[\bar{a} \bar{b} \bar{c}] = 4X$.
Substituting this into the volume equation: $|4X| = 72$.
Therefore,$4X = 72$,which gives $X = \frac{72}{4} = 18$.
Thus,$|\bar{b} \times \bar{c}| = 18$.

(A) The volume $V$ of a parallelepiped with coterminous edges represented by vectors $\vec{a}$,$\vec{b}$,and $\vec{c}$ is given by the absolute value of the scalar triple product: $V = |\vec{a} \cdot (\vec{b} \times \vec{c})|$.
This is equivalent to the determinant of the matrix formed by the components of the vectors:
$V = |\det \begin{bmatrix} 1 & 0 & -1 \\ \lambda & 1 & 1-\lambda \\ \mu & \lambda & 1+\lambda-\mu \end{bmatrix}|$.
Expanding the determinant along the first row:
$V = |1(1(1+\lambda-\mu) - \lambda(1-\lambda)) - 0 + (-1)(\lambda(\lambda) - 1(\mu))|$.
$V = |(1+\lambda-\mu - \lambda + \lambda^2) - (\lambda^2 - \mu)|$.
$V = |1 + \lambda^2 - \mu - \lambda - \lambda^2 + \mu|$.
$V = |1 - \lambda|$.
The volume $V = |1 - \lambda|$ depends only on $\lambda$ and is independent of $\mu$.

(A) Given that $\bar{a}$ and $\bar{c}$ are unit vectors,so $|\bar{a}| = 1$ and $|\bar{c}| = 1$. The angle between them is $\theta = \frac{\pi}{3}$,so $\bar{a} \cdot \bar{c} = |\bar{a}| |\bar{c}| \cos(\frac{\pi}{3}) = 1 \times 1 \times \frac{1}{2} = \frac{1}{2}$.
Using the vector triple product formula,$\bar{a} \times (\bar{b} \times \bar{c}) = (\bar{a} \cdot \bar{c})\bar{b} - (\bar{a} \cdot \bar{b})\bar{c}$.
Substitute this into the given equation: $((\bar{a} \cdot \bar{c})\bar{b} - (\bar{a} \cdot \bar{b})\bar{c}) \cdot (\bar{a} \times \bar{c}) = 5$.
Since $(\bar{a} \times \bar{c})$ is perpendicular to both $\bar{a}$ and $\bar{c}$,we have $\bar{c} \cdot (\bar{a} \times \bar{c}) = 0$.
Thus,the equation simplifies to $(\bar{a} \cdot \bar{c}) \bar{b} \cdot (\bar{a} \times \bar{c}) = 5$.
Substituting $\bar{a} \cdot \bar{c} = \frac{1}{2}$,we get $\frac{1}{2} [\bar{b} \bar{a} \bar{c}] = 5$,which implies $[\bar{b} \bar{a} \bar{c}] = 10$.
Since the scalar triple product is cyclic,$[\bar{b} \bar{a} \bar{c}] = [\bar{a} \bar{b} \bar{c}] = 10$.

(B) The volume $V$ of a parallelepiped with coterminous edges $\vec{a}, \vec{b}, \vec{c}$ is given by the scalar triple product $|\vec{a} \cdot (\vec{b} \times \vec{c})|$.
Given $\vec{a} = \hat{i}+x \hat{j}+\hat{k}$,$\vec{b} = \hat{j}+x \hat{k}$,and $\vec{c} = x \hat{i}+\hat{k}$.
The scalar triple product is the determinant:
$V = |\det \begin{bmatrix} 1 & x & 1 \\ 0 & 1 & x \\ x & 0 & 1 \end{bmatrix}|$
Expanding along the first row:
$V = |1(1-0) - x(0-x^2) + 1(0-x)| = |1 + x^3 - x|$.
Let $f(x) = x^3 - x + 1$. To find the extrema,we set $f'(x) = 3x^2 - 1 = 0$,which gives $x^2 = \frac{1}{3}$,so $x = \pm \frac{1}{\sqrt{3}}$.
At $x = \frac{1}{\sqrt{3}}$,$f(\frac{1}{\sqrt{3}}) = (\frac{1}{\sqrt{3}})^3 - \frac{1}{\sqrt{3}} + 1 = \frac{1}{3\sqrt{3}} - \frac{3}{3\sqrt{3}} + 1 = 1 - \frac{2}{3\sqrt{3}}$.
At $x = -\frac{1}{\sqrt{3}}$,$f(-\frac{1}{\sqrt{3}}) = -\frac{1}{3\sqrt{3}} + \frac{1}{\sqrt{3}} + 1 = 1 + \frac{2}{3\sqrt{3}}$.
Since volume must be non-negative,the maximum value is $1 + \frac{2}{3\sqrt{3}}$ and the minimum value is $1 - \frac{2}{3\sqrt{3}}$.

(C) The volume of a tetrahedron with coterminus edges $\bar{a}$,$\bar{b}$,and $\bar{c}$ is given by the formula: $V = \frac{1}{6} |[\bar{a} \bar{b} \bar{c}]|$.
Given $V = 570$,we have $|[\bar{a} \bar{b} \bar{c}]| = 6 \times 570 = 3420$.
The scalar triple product $[\bar{a} \bar{b} \bar{c}]$ is the determinant of the matrix formed by the components of the vectors:
$[\bar{a} \bar{b} \bar{c}] = \begin{vmatrix} -12 & 0 & p \\ 0 & 3 & -1 \\ 2 & 1 & -15 \end{vmatrix}$.
Expanding the determinant along the first row:
$[\bar{a} \bar{b} \bar{c}] = -12(3(-15) - (-1)(1)) - 0 + p(0(1) - 3(2))$
$= -12(-45 + 1) + p(-6)$
$= -12(-44) - 6p = 528 - 6p$.
Since $|528 - 6p| = 3420$,we have two cases:
Case $1$: $528 - 6p = 3420 \implies -6p = 2892 \implies p = -482$.
Case $2$: $528 - 6p = -3420 \implies -6p = -3948 \implies p = 658$.
Comparing with the given options,$p = -482$ is the correct value.

(C) The volume of a tetrahedron with coterminus edges $\bar{a}, \bar{b}, \bar{c}$ is given by $V_{tetra} = \frac{1}{6} |[\bar{a} \bar{b} \bar{c}]|$.
Given $V_{tetra} = \frac{64}{3}$,we have $\frac{1}{6} |[\bar{a} \bar{b} \bar{c}]| = \frac{64}{3}$,which implies $|[\bar{a} \bar{b} \bar{c}]| = 128$.
The volume of a parallelepiped with coterminus edges $\bar{a}+\bar{b}, \bar{b}+\bar{c}, \bar{c}+\bar{a}$ is given by the scalar triple product $|[(\bar{a}+\bar{b}) (\bar{b}+\bar{c}) (\bar{c}+\bar{a})]|$.
Using the property of scalar triple products,$[(\bar{a}+\bar{b}) (\bar{b}+\bar{c}) (\bar{c}+\bar{a})] = 2 [\bar{a} \bar{b} \bar{c}]$.
Thus,the volume is $|2 [\bar{a} \bar{b} \bar{c}]| = 2 \times 128 = 256$ cubic units.

(NONE) The volume of a parallelepiped is given by the scalar triple product $|[\bar{a} \bar{b} \bar{c}]|$.
First,calculate the scalar triple product:
$[\bar{a} \bar{b} \bar{c}] = \begin{vmatrix} 1 & 1 & 1 \\ 2 & 4 & -1 \\ 1 & 1 & 3 \end{vmatrix} = 1(12 - (-1)) - 1(6 - (-1)) + 1(2 - 4) = 1(13) - 1(7) + 1(-2) = 13 - 7 - 2 = 4$.
So,the volume $V = 4$.
The area of the base formed by $\bar{a}$ and $\bar{b}$ is $|\bar{a} \times \bar{b}|$.
$\bar{a} \times \bar{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & 4 & -1 \end{vmatrix} = \hat{i}(-1 - 4) - \hat{j}(-1 - 2) + \hat{k}(4 - 2) = -5\hat{i} + 3\hat{j} + 2\hat{k}$.
Area of base = $|-5\hat{i} + 3\hat{j} + 2\hat{k}| = \sqrt{(-5)^2 + 3^2 + 2^2} = \sqrt{25 + 9 + 4} = \sqrt{38}$.
The altitude $h$ is given by $V / \text{Area of base} = 4 / \sqrt{38} = 4\sqrt{38}/38 = 2\sqrt{38}/19$.

(B) Given the determinant equation:
$\left|\begin{array}{lll} a & a^2 & 1+a^3 \\ b & b^2 & 1+b^3 \\ c & c^2 & 1+c^3 \end{array}\right|=0$
This can be split into two determinants:
$\left|\begin{array}{lll} a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1 \end{array}\right| + \left|\begin{array}{lll} a & a^2 & a^3 \\ b & b^2 & b^3 \\ c & c^2 & c^3 \end{array}\right| = 0$
Taking $abc$ common from the second determinant:
$\left|\begin{array}{lll} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array}\right| + abc \left|\begin{array}{lll} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array}\right| = 0$
$(1 + abc) \left|\begin{array}{lll} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array}\right| = 0$
The determinant $\left|\begin{array}{lll} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array}\right|$ is the Vandermonde determinant,which equals $(a-b)(b-c)(c-a)$.
Since the vectors $\overline{p}, \overline{q}, \overline{r}$ are non-coplanar,their scalar triple product is non-zero,which implies $a, b, c$ are distinct,so $(a-b)(b-c)(c-a) \neq 0$.
Therefore,$1 + abc = 0$,which gives $abc = -1$.

(D) The volume $V$ of the parallelepiped formed by vectors $\vec{a}, \vec{b}, \vec{c}$ is given by the scalar triple product $|[\vec{a} \vec{b} \vec{c}]|$.
$V = \begin{vmatrix} 1 & m & 1 \\ 0 & 1 & m \\ m & 0 & 1 \end{vmatrix} = 1(1 - 0) - m(0 - m^2) + 1(0 - m) = 1 + m^3 - m$.
To find the minimum volume,we differentiate $V$ with respect to $m$:
$\frac{dV}{dm} = 3m^2 - 1$.
Setting $\frac{dV}{dm} = 0$,we get $3m^2 = 1$,which implies $m^2 = \frac{1}{3}$,so $m = \pm \frac{1}{\sqrt{3}}$.
Since we are looking for the volume (which must be positive),we consider the magnitude. For $m > 0$,we take $m = \frac{1}{\sqrt{3}}$.
Checking the second derivative: $\frac{d^2V}{dm^2} = 6m$.
For $m = \frac{1}{\sqrt{3}}$,$\frac{d^2V}{dm^2} = 6(\frac{1}{\sqrt{3}}) > 0$,which confirms that the volume is minimum at $m = \frac{1}{\sqrt{3}}$.

(A) Given expression: $\overline{a} \cdot [(\overline{b} + \overline{c}) \times (\overline{a} + \overline{b} + \overline{c})]$
Using the distributive property of the cross product:
$(\overline{b} + \overline{c}) \times (\overline{a} + \overline{b} + \overline{c}) = \overline{b} \times \overline{a} + \overline{b} \times \overline{b} + \overline{b} \times \overline{c} + \overline{c} \times \overline{a} + \overline{c} \times \overline{b} + \overline{c} \times \overline{c}$
Since $\overline{x} \times \overline{x} = 0$ and $\overline{c} \times \overline{b} = -(\overline{b} \times \overline{c})$,we have:
$= \overline{b} \times \overline{a} + 0 + \overline{b} \times \overline{c} + \overline{c} \times \overline{a} - (\overline{b} \times \overline{c}) + 0 = \overline{b} \times \overline{a} + \overline{c} \times \overline{a}$
Now,taking the dot product with $\overline{a}$:
$\overline{a} \cdot (\overline{b} \times \overline{a} + \overline{c} \times \overline{a}) = \overline{a} \cdot (\overline{b} \times \overline{a}) + \overline{a} \cdot (\overline{c} \times \overline{a})$
$= [\overline{a} \overline{b} \overline{a}] + [\overline{a} \overline{c} \overline{a}]$
Since the scalar triple product is zero if any two vectors are identical:
$= 0 + 0 = 0$

(D) Given that $|\overline{a}| = |\overline{c}| = 1$ and the angle between $\overline{a}$ and $\overline{c}$ is $\frac{\pi}{3}$.
$\overline{a} \cdot \overline{c} = |\overline{a}||\overline{c}| \cos \frac{\pi}{3} = 1 \times 1 \times \frac{1}{2} = \frac{1}{2}$.
Using the vector triple product formula $\overline{a} \times (\overline{b} \times \overline{c}) = (\overline{a} \cdot \overline{c}) \overline{b} - (\overline{a} \cdot \overline{b}) \overline{c}$.
Substituting this into the given equation: $((\overline{a} \cdot \overline{c}) \overline{b} - (\overline{a} \cdot \overline{b}) \overline{c}) \cdot (\overline{a} \times \overline{c}) = 5$.
Since $(\overline{c} \cdot (\overline{a} \times \overline{c})) = 0$ (as the scalar triple product with two identical vectors is zero),the equation simplifies to:
$(\overline{a} \cdot \overline{c}) \overline{b} \cdot (\overline{a} \times \overline{c}) = 5$.
$\frac{1}{2} [\overline{b} \overline{a} \overline{c}] = 5 \Rightarrow [\overline{b} \overline{a} \overline{c}] = 10$.
Using the property of scalar triple products,$[\overline{b} \overline{a} \overline{c}] = -[\overline{a} \overline{b} \overline{c}]$.
Therefore,$-[\overline{a} \overline{b} \overline{c}] = 10 \Rightarrow [\overline{a} \overline{b} \overline{c}] = -10$.
Finally,$5[\overline{a} \overline{b} \overline{c}] = 5 \times (-10) = -50$.

(C) The scalar triple product $\overline{a} \cdot(\overline{b} \times \overline{c})$ is given by the determinant of the components of the vectors $\overline{a}, \overline{b},$ and $\overline{c}$.
$[\overline{a} \overline{b} \overline{c}] = \begin{vmatrix} 1 & 0 & -1 \\ x & 1 & 1-x \\ y & x & 1+x-y \end{vmatrix}$
Applying the column operation $C_3 \rightarrow C_3 + C_1$:
$[\overline{a} \overline{b} \overline{c}] = \begin{vmatrix} 1 & 0 & -1+1 \\ x & 1 & 1-x+x \\ y & x & 1+x-y+y \end{vmatrix} = \begin{vmatrix} 1 & 0 & 0 \\ x & 1 & 1 \\ y & x & 1+x \end{vmatrix}$
Expanding along the first row:
$= 1 \times \begin{vmatrix} 1 & 1 \\ x & 1+x \end{vmatrix} - 0 + 0$
$= 1 \times ((1+x) - x) = 1 \times 1 = 1$
Since the result is $1$,which is a constant independent of $x$ and $y$,the value of $\overline{a} \cdot(\overline{b} \times \overline{c})$ depends on neither $x$ nor $y$.

(C) Given that $\overline{a}, \overline{b}, \overline{c}$ are mutually perpendicular vectors,we have $\overline{a} \cdot \overline{b} = 0, \overline{b} \cdot \overline{c} = 0, \overline{c} \cdot \overline{a} = 0$ and magnitudes $|\overline{a}|=1, |\overline{b}|=2, |\overline{c}|=3$.
The scalar triple product is defined as $[\overline{a}+\overline{b}+\overline{c} \quad \overline{b}-\overline{a} \quad \overline{c}] = (\overline{a}+\overline{b}+\overline{c}) \cdot ((\overline{b}-\overline{a}) \times \overline{c})$.
Expanding the cross product: $(\overline{b}-\overline{a}) \times \overline{c} = \overline{b} \times \overline{c} - \overline{a} \times \overline{c}$.
Now,substitute this back: $(\overline{a}+\overline{b}+\overline{c}) \cdot (\overline{b} \times \overline{c} - \overline{a} \times \overline{c})$.
Since $\overline{a}, \overline{b}, \overline{c}$ are mutually perpendicular,$\overline{a} \cdot (\overline{b} \times \overline{c}) = |\overline{a}| |\overline{b}| |\overline{c}| = 1 \times 2 \times 3 = 6$.
Also,$\overline{b} \cdot (\overline{b} \times \overline{c}) = 0$ and $\overline{c} \cdot (\overline{b} \times \overline{c}) = 0$.
Similarly,$\overline{a} \cdot (\overline{a} \times \overline{c}) = 0, \overline{b} \cdot (\overline{a} \times \overline{c}) = 0, \overline{c} \cdot (\overline{a} \times \overline{c}) = 0$.
The expression simplifies to: $\overline{a} \cdot (\overline{b} \times \overline{c}) - \overline{a} \cdot (\overline{a} \times \overline{c}) + \overline{b} \cdot (\overline{b} \times \overline{c}) - \overline{b} \cdot (\overline{a} \times \overline{c}) + \overline{c} \cdot (\overline{b} \times \overline{c}) - \overline{c} \cdot (\overline{a} \times \overline{c})$.
This reduces to $\overline{a} \cdot (\overline{b} \times \overline{c}) - (-\overline{b} \cdot (\overline{a} \times \overline{c})) = [\overline{a} \overline{b} \overline{c}] + [\overline{b} \overline{a} \overline{c}] = [\overline{a} \overline{b} \overline{c}] + [\overline{a} \overline{b} \overline{c}] = 2[\overline{a} \overline{b} \overline{c}]$.
Since $[\overline{a} \overline{b} \overline{c}] = 1 \times 2 \times 3 = 6$,the final value is $2 \times 6 = 12$.

(A) The volume of a tetrahedron with vertices $A, B, C, D$ is given by $V = \frac{1}{6} |[\vec{AB}, \vec{AC}, \vec{AD}]|$.
Given vertices: $A(1, -6, 10)$,$B(-1, -3, 7)$,$C(5, -1, k)$,$D(7, -4, 7)$.
Vectors are:
$\vec{AB} = (-1-1)\hat{i} + (-3-(-6))\hat{j} + (7-10)\hat{k} = -2\hat{i} + 3\hat{j} - 3\hat{k}$
$\vec{AC} = (5-1)\hat{i} + (-1-(-6))\hat{j} + (k-10)\hat{k} = 4\hat{i} + 5\hat{j} + (k-10)\hat{k}$
$\vec{AD} = (7-1)\hat{i} + (-4-(-6))\hat{j} + (7-10)\hat{k} = 6\hat{i} + 2\hat{j} - 3\hat{k}$
Volume $= \frac{1}{6} |\det(\vec{AB}, \vec{AC}, \vec{AD})| = 11$.
$\det = \begin{vmatrix} -2 & 3 & -3 \\ 4 & 5 & k-10 \\ 6 & 2 & -3 \end{vmatrix} = -2(-15 - 2(k-10)) - 3(-12 - 6(k-10)) - 3(8 - 30)$
$= -2(-15 - 2k + 20) - 3(-12 - 6k + 60) - 3(-22)$
$= -2(5 - 2k) - 3(48 - 6k) + 66$
$= -10 + 4k - 144 + 18k + 66 = 22k - 88$.
Since $V = 11$,$\frac{1}{6} |22k - 88| = 11 \Rightarrow |22k - 88| = 66$.
Case $1$: $22k - 88 = 66 \Rightarrow 22k = 154 \Rightarrow k = 7$.
Case $2$: $22k - 88 = -66 \Rightarrow 22k = 22 \Rightarrow k = 1$.
Given the options,$k=7$ is the correct value.

(A) Let the scalar triple product be denoted by $[\bar{a} \bar{b} \bar{c}] = \bar{a} \cdot (\bar{b} \times \bar{c})$.
We need to evaluate $(\bar{u}+\bar{v}-\bar{w}) \cdot [(\bar{u}-\bar{v}) \times (\bar{v}-\bar{w})]$.
First,expand 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 $\bar{u} \times \bar{v} - \bar{u} \times \bar{w} + \bar{v} \times \bar{w}$.
Now,take the dot product with $(\bar{u}+\bar{v}-\bar{w})$:
$(\bar{u}+\bar{v}-\bar{w}) \cdot (\bar{u} \times \bar{v} - \bar{u} \times \bar{w} + \bar{v} \times \bar{w})$
$= \bar{u} \cdot (\bar{u} \times \bar{v}) - \bar{u} \cdot (\bar{u} \times \bar{w}) + \bar{u} \cdot (\bar{v} \times \bar{w}) + \bar{v} \cdot (\bar{u} \times \bar{v}) - \bar{v} \cdot (\bar{u} \times \bar{w}) + \bar{v} \cdot (\bar{v} \times \bar{w}) - \bar{w} \cdot (\bar{u} \times \bar{v}) + \bar{w} \cdot (\bar{u} \times \bar{w}) - \bar{w} \cdot (\bar{v} \times \bar{w})$.
Using the property that the scalar triple product is zero if any two vectors are identical:
$= 0 - 0 + [\bar{u} \bar{v} \bar{w}] + 0 - [\bar{v} \bar{u} \bar{w}] + 0 - [\bar{w} \bar{u} \bar{v}] + 0 - 0$.
$= [\bar{u} \bar{v} \bar{w}] + [\bar{u} \bar{v} \bar{w}] - [\bar{u} \bar{v} \bar{w}] = [\bar{u} \bar{v} \bar{w}] = \bar{u} \cdot (\bar{v} \times \bar{w})$.

(C) We know that the scalar triple product is defined as $[\bar{x} \quad \bar{y} \quad \bar{z}] = (\bar{x} \times \bar{y}) \cdot \bar{z}$.
Given expression: $[\bar{a} \times \bar{b} \quad \bar{b} \times \bar{c} \quad \bar{c} \times \bar{a}] = ((\bar{a} \times \bar{b}) \times (\bar{b} \times \bar{c})) \cdot (\bar{c} \times \bar{a})$.
Using the vector identity $(\bar{u} \times \bar{v}) \times \bar{w} = (\bar{u} \cdot \bar{w})\bar{v} - (\bar{v} \cdot \bar{w})\bar{u}$,we have:
$(\bar{a} \times \bar{b}) \times (\bar{b} \times \bar{c}) = [\bar{a} \quad \bar{b} \quad \bar{c}]\bar{b}$.
Substituting this into the expression:
$[\bar{a} \times \bar{b} \quad \bar{b} \times \bar{c} \quad \bar{c} \times \bar{a}] = ([\bar{a} \quad \bar{b} \quad \bar{c}]\bar{b}) \cdot (\bar{c} \times \bar{a})$
$= [\bar{a} \quad \bar{b} \quad \bar{c}] (\bar{b} \cdot (\bar{c} \times \bar{a}))$
$= [\bar{a} \quad \bar{b} \quad \bar{c}] [\bar{b} \quad \bar{c} \quad \bar{a}]$
Since the scalar triple product is cyclic,$[\bar{b} \quad \bar{c} \quad \bar{a}] = [\bar{a} \quad \bar{b} \quad \bar{c}]$.
Therefore,$[\bar{a} \times \bar{b} \quad \bar{b} \times \bar{c} \quad \bar{c} \times \bar{a}] = [\bar{a} \quad \bar{b} \quad \bar{c}]^2$.
Comparing this with $\lambda [\bar{a} \quad \bar{b} \quad \bar{c}]^2$,we get $\lambda = 1$.

(A) Since $\overline{a}, \overline{b}, \overline{c}$ are coplanar vectors,their scalar triple product is zero,i.e.,$[\overline{a}, \overline{b}, \overline{c}] = 0$.
Let $\overline{\alpha} = 2 \overline{a} - \overline{b}$,$\overline{\beta} = 2 \overline{b} - \overline{c}$,and $\overline{\gamma} = 2 \overline{c} - \overline{a}$.
The scalar triple product $[\overline{\alpha}, \overline{\beta}, \overline{\gamma}]$ can be expressed as the determinant of the coefficients multiplied by $[\overline{a}, \overline{b}, \overline{c}]$:
$[\overline{\alpha}, \overline{\beta}, \overline{\gamma}] = \begin{vmatrix} 2 & -1 & 0 \\ 0 & 2 & -1 \\ -1 & 0 & 2 \end{vmatrix} [\overline{a}, \overline{b}, \overline{c}]$.
Calculating the determinant: $2(4 - 0) - (-1)(0 - 1) + 0 = 2(4) + 1(-1) = 8 - 1 = 7$.
Thus,$[\overline{\alpha}, \overline{\beta}, \overline{\gamma}] = 7 \times [\overline{a}, \overline{b}, \overline{c}] = 7 \times 0 = 0$.

(B) First,note that the determinant on the right side is equal to the scalar triple product $[\bar{a} \quad \bar{b} \quad \bar{c}]$.
$\begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix} = [\bar{a} \quad \bar{b} \quad \bar{c}]$.
Now,calculate the scalar triple product on the left side:
$[3 \bar{a}+\bar{b} \quad 3 \bar{b}+\bar{c} \quad 3 \bar{c}+\bar{a}] = (3 \bar{a}+\bar{b}) \cdot ((3 \bar{b}+\bar{c}) \times (3 \bar{c}+\bar{a}))$.
Expanding the cross product:
$(3 \bar{b}+\bar{c}) \times (3 \bar{c}+\bar{a}) = 9(\bar{b} \times \bar{c}) + 3(\bar{b} \times \bar{a}) + 3(\bar{c} \times \bar{c}) + (\bar{c} \times \bar{a}) = 9(\bar{b} \times \bar{c}) + 3(\bar{b} \times \bar{a}) + (\bar{c} \times \bar{a})$.
Now take the dot product with $(3 \bar{a}+\bar{b})$:
$= (3 \bar{a}+\bar{b}) \cdot (9(\bar{b} \times \bar{c}) + 3(\bar{b} \times \bar{a}) + (\bar{c} \times \bar{a}))$
$= 27[\bar{a} \quad \bar{b} \quad \bar{c}] + 9[\bar{a} \quad \bar{b} \quad \bar{a}] + 3[\bar{a} \quad \bar{c} \quad \bar{a}] + 9[\bar{b} \quad \bar{b} \quad \bar{c}] + 3[\bar{b} \quad \bar{b} \quad \bar{a}] + [\bar{b} \quad \bar{c} \quad \bar{a}]$.
Since any scalar triple product with two identical vectors is $0$,this simplifies to:
$= 27[\bar{a} \quad \bar{b} \quad \bar{c}] + [\bar{b} \quad \bar{c} \quad \bar{a}] = 27[\bar{a} \quad \bar{b} \quad \bar{c}] + [\bar{a} \quad \bar{b} \quad \bar{c}] = 28[\bar{a} \quad \bar{b} \quad \bar{c}]$.
Comparing this with $\lambda [\bar{a} \quad \bar{b} \quad \bar{c}]$,we get $\lambda = 28$.

(D) The scalar triple product is defined as $\left[\bar{a} \bar{b} \bar{c}\right] = \bar{a} \cdot (\bar{b} \times \bar{c})$.
Since $\bar{a}$ is perpendicular to both $\bar{b}$ and $\bar{c}$,$\bar{a}$ must be parallel to the vector $\bar{b} \times \bar{c}$.
Let $\hat{n}$ be the unit vector perpendicular to $\bar{b}$ and $\bar{c}$. Then $\bar{b} \times \bar{c} = |\bar{b}||\bar{c}| \sin\left(\frac{\pi}{3}\right) \hat{n} = 3 \times 4 \times \frac{\sqrt{3}}{2} \hat{n} = 6\sqrt{3} \hat{n}$.
Since $\bar{a}$ is perpendicular to $\bar{b}$ and $\bar{c}$,$\bar{a} = |\bar{a}| \hat{n} = 2 \hat{n}$.
Therefore,$\left[\bar{a} \bar{b} \bar{c}\right] = \bar{a} \cdot (\bar{b} \times \bar{c}) = (2 \hat{n}) \cdot (6\sqrt{3} \hat{n}) = 12\sqrt{3} (\hat{n} \cdot \hat{n}) = 12\sqrt{3} \times 1 = 12\sqrt{3}$.

(B) The volume of a parallelepiped is given by the scalar triple product $|[\bar{u} \bar{v} \bar{w}]| = 1$.
Calculating the determinant:
$\begin{vmatrix} 1 & 1 & \lambda \\ 1 & 1 & 3 \\ 2 & 1 & 1 \end{vmatrix} = \pm 1$.
Expanding along the first row:
$1(1-3) - 1(1-6) + \lambda(1-2) = \pm 1$
$-2 + 5 - \lambda = \pm 1$
$3 - \lambda = \pm 1$.
Case $1$: $3 - \lambda = 1 \Rightarrow \lambda = 2$.
Case $2$: $3 - \lambda = -1 \Rightarrow \lambda = 4$.
Given $\bar{u} = \hat{i} + \hat{j} + \lambda \hat{k}$ and $\bar{w} = 2\hat{i} + \hat{j} + \hat{k}$.
For $\lambda = 2$: $\bar{u} = \hat{i} + \hat{j} + 2\hat{k}$.
$\cos \theta = \frac{\bar{u} \cdot \bar{w}}{|\bar{u}| |\bar{w}|} = \frac{(1)(2) + (1)(1) + (2)(1)}{\sqrt{1^2+1^2+2^2} \sqrt{2^2+1^2+1^2}} = \frac{2+1+2}{\sqrt{6} \cdot \sqrt{6}} = \frac{5}{6}$.

(C) Let the scalar triple product be denoted by $[\overline{a} \ \overline{b} \ \overline{c}] = V$.
The given expression is $[(\overline{a}+2 \overline{b}+3 \overline{c}) \times(\overline{b}+2 \overline{c}+3 \overline{a})] \cdot(\overline{c}+2 \overline{a}+3 \overline{b}) = 54$.
This is the scalar triple product of the vectors $\overline{u} = \overline{a}+2 \overline{b}+3 \overline{c}$,$\overline{v} = \overline{b}+2 \overline{c}+3 \overline{a}$,and $\overline{w} = \overline{c}+2 \overline{a}+3 \overline{b}$.
The scalar triple product can be represented as the determinant of the coefficients:
$\begin{vmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \\ 2 & 3 & 1 \end{vmatrix} [\overline{a} \ \overline{b} \ \overline{c}] = 54$.
Calculating the determinant:
$1(1-6) - 2(3-4) + 3(9-2) = 1(-5) - 2(-1) + 3(7) = -5 + 2 + 21 = 18$.
Thus,$18 [\overline{a} \ \overline{b} \ \overline{c}] = 54$.
Therefore,$[\overline{a} \ \overline{b} \ \overline{c}] = \frac{54}{18} = 3$.

(A) The volume $V$ of a parallelepiped formed by vectors $\vec{a}, \vec{b}, \vec{c}$ is given by the scalar triple product $|[\vec{a} \vec{b} \vec{c}]|$.
$V = \left| \begin{vmatrix} 1 & \alpha & 1 \\ 0 & 1 & \alpha \\ \alpha & 0 & 1 \end{vmatrix} \right| = |1(1-0) - \alpha(0-\alpha^2) + 1(0-\alpha)| = |1 + \alpha^3 - \alpha|$.
Let $f(\alpha) = 1 + \alpha^3 - \alpha$.
To find the maximum,we differentiate $f(\alpha)$ with respect to $\alpha$:
$f'(\alpha) = 3\alpha^2 - 1$.
Setting $f'(\alpha) = 0$,we get $3\alpha^2 = 1$,which implies $\alpha = \pm \frac{1}{\sqrt{3}}$.
Now,we check the second derivative: $f''(\alpha) = 6\alpha$.
For $\alpha = \frac{1}{\sqrt{3}}$,$f''(\alpha) = 6(\frac{1}{\sqrt{3}}) > 0$ (local minimum).
For $\alpha = -\frac{1}{\sqrt{3}}$,$f''(\alpha) = 6(-\frac{1}{\sqrt{3}}) < 0$ (local maximum).
Thus,the volume is maximum at $\alpha = -\frac{1}{\sqrt{3}}$.

(C) Let the expression be $E = (\bar{a}+2 \bar{b}+\bar{c}) \cdot [(\bar{a}-\bar{b}) \times (\bar{a}-\bar{b}-\bar{c})]$.
First,simplify the cross product part:
$(\bar{a}-\bar{b}) \times (\bar{a}-\bar{b}-\bar{c}) = (\bar{a} \times \bar{a}) - (\bar{a} \times \bar{b}) - (\bar{a} \times \bar{c}) - (\bar{b} \times \bar{a}) + (\bar{b} \times \bar{b}) + (\bar{b} \times \bar{c})$.
Since $\bar{a} \times \bar{a} = 0$,$\bar{b} \times \bar{b} = 0$,and $\bar{b} \times \bar{a} = -(\bar{a} \times \bar{b})$,we get:
$= 0 - (\bar{a} \times \bar{b}) - (\bar{a} \times \bar{c}) + (\bar{a} \times \bar{b}) + 0 + (\bar{b} \times \bar{c}) = (\bar{b} \times \bar{c}) - (\bar{a} \times \bar{c}) = (\bar{b} \times \bar{c}) + (\bar{c} \times \bar{a})$.
Now,compute the dot product:
$E = (\bar{a}+2 \bar{b}+\bar{c}) \cdot (\bar{b} \times \bar{c} + \bar{c} \times \bar{a})$.
$= \bar{a} \cdot (\bar{b} \times \bar{c}) + \bar{a} \cdot (\bar{c} \times \bar{a}) + 2\bar{b} \cdot (\bar{b} \times \bar{c}) + 2\bar{b} \cdot (\bar{c} \times \bar{a}) + \bar{c} \cdot (\bar{b} \times \bar{c}) + \bar{c} \cdot (\bar{c} \times \bar{a})$.
Using properties of scalar triple products,terms like $\bar{a} \cdot (\bar{c} \times \bar{a}) = 0$ and $\bar{b} \cdot (\bar{b} \times \bar{c}) = 0$ vanish.
$E = [\bar{a} \bar{b} \bar{c}] + 0 + 0 + 2[\bar{b} \bar{c} \bar{a}] + 0 + 0$.
Since $[\bar{b} \bar{c} \bar{a}] = [\bar{a} \bar{b} \bar{c}]$,we have:
$E = [\bar{a} \bar{b} \bar{c}] + 2[\bar{a} \bar{b} \bar{c}] = 3[\bar{a} \bar{b} \bar{c}]$.

(C) The volume $V$ of the parallelepiped formed by vectors $\vec{a} = \hat{i}+\alpha \hat{j}+\hat{k}$,$\vec{b} = \hat{j}+\alpha \hat{k}$,and $\vec{c} = \alpha \hat{i}+\hat{k}$ is given by the absolute value of the scalar triple product:
$V = |\vec{a} \cdot (\vec{b} \times \vec{c})| = \left|\begin{array}{ccc} 1 & \alpha & 1 \\ 0 & 1 & \alpha \\ \alpha & 0 & 1 \end{array}\right|$
Expanding the determinant:
$V = 1(1 - 0) - \alpha(0 - \alpha^2) + 1(0 - \alpha) = 1 + \alpha^3 - \alpha$
To find the minimum volume,we differentiate $V$ with respect to $\alpha$ and set it to zero:
$\frac{dV}{d\alpha} = 3\alpha^2 - 1 = 0$
$\alpha^2 = \frac{1}{3} \implies \alpha = \pm \frac{1}{\sqrt{3}}$
Now,we check the second derivative:
$\frac{d^2V}{d\alpha^2} = 6\alpha$
For $\alpha = \frac{1}{\sqrt{3}}$,$\frac{d^2V}{d\alpha^2} = 6(\frac{1}{\sqrt{3}}) > 0$,which indicates a local minimum.
Thus,the volume is minimum at $\alpha = \frac{1}{\sqrt{3}}$.

(C) Let the position vectors of the vertices be $\vec{a} = \hat{i}-6 \hat{j}+10 \hat{k}$,$\vec{b} = -\hat{i}-3 \hat{j}+7 \hat{k}$,$\vec{c} = 5 \hat{i}-\hat{j}+\lambda \hat{k}$,and $\vec{d} = 7 \hat{i}-4 \hat{j}+7 \hat{k}$.
The vectors representing the edges are:
$\vec{AB} = \vec{b} - \vec{a} = (-1-1)\hat{i} + (-3+6)\hat{j} + (7-10)\hat{k} = -2\hat{i} + 3\hat{j} - 3\hat{k}$
$\vec{AC} = \vec{c} - \vec{a} = (5-1)\hat{i} + (-1+6)\hat{j} + (\lambda-10)\hat{k} = 4\hat{i} + 5\hat{j} + (\lambda-10)\hat{k}$
$\vec{AD} = \vec{d} - \vec{a} = (7-1)\hat{i} + (-4+6)\hat{j} + (7-10)\hat{k} = 6\hat{i} + 2\hat{j} - 3\hat{k}$
The volume of the tetrahedron is given by $V = \frac{1}{6} |[\vec{AB} \ \vec{AC} \ \vec{AD}]|$.
Given $V = 11$,we have $11 = \frac{1}{6} |\det(\vec{AB}, \vec{AC}, \vec{AD})|$.
$66 = |\det \begin{bmatrix} -2 & 3 & -3 \\ 4 & 5 & \lambda-10 \\ 6 & 2 & -3 \end{bmatrix}|$.
Calculating the determinant:
$-2(5(-3) - 2(\lambda-10)) - 3(4(-3) - 6(\lambda-10)) - 3(4(2) - 6(5))$
$= -2(-15 - 2\lambda + 20) - 3(-12 - 6\lambda + 60) - 3(8 - 30)$
$= -2(5 - 2\lambda) - 3(48 - 6\lambda) - 3(-22)$
$= -10 + 4\lambda - 144 + 18\lambda + 66$
$= 22\lambda - 88$.
Since $66 = |22\lambda - 88|$,we have $22\lambda - 88 = 66$ or $22\lambda - 88 = -66$.
Case $1$: $22\lambda = 154 \Rightarrow \lambda = 7$.
Case $2$: $22\lambda = 22 \Rightarrow \lambda = 1$.
Given the options,$\lambda = 7$ is the correct value.

(A) The volume of a parallelepiped with coterminous edges $\bar{a}, \bar{b}, \bar{c}$ is given by the scalar triple product $|[\bar{a} \bar{b} \bar{c}]|$.
Given,volume $= 158$.
$|\bar{a} \cdot (\bar{b} \times \bar{c})| = \left|\begin{vmatrix} 1 & 1 & n \\ 2 & 4 & -n \\ 1 & n & 3 \end{vmatrix}\right| = 158$.
Expanding the determinant:
$1(12 - (-n^2)) - 1(6 - (-n)) + n(2n - 4) = \pm 158$.
$(12 + n^2) - (6 + n) + (2n^2 - 4n) = \pm 158$.
$3n^2 - 5n + 6 = \pm 158$.
Case $1$: $3n^2 - 5n + 6 = 158 \Rightarrow 3n^2 - 5n - 152 = 0$.
Using the quadratic formula $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$n = \frac{5 \pm \sqrt{25 - 4(3)(-152)}}{6} = \frac{5 \pm \sqrt{25 + 1824}}{6} = \frac{5 \pm \sqrt{1849}}{6} = \frac{5 \pm 43}{6}$.
Since $n \geq 0$,$n = \frac{48}{6} = 8$.
Case $2$: $3n^2 - 5n + 6 = -158 \Rightarrow 3n^2 - 5n + 164 = 0$.
The discriminant $D = 25 - 4(3)(164) < 0$,so no real solution exists.
Thus,$n = 8$.

(C) The volume of a tetrahedron with vertices $A, B, C, D$ is given by $V = \frac{1}{6} |(\vec{AB} \times \vec{AC}) \cdot \vec{AD}| = \frac{1}{6} |[\vec{AB} \vec{AC} \vec{AD}]|$.
First,we find the vectors:
$\vec{AB} = (-3-1)\hat{i} + (-1-2)\hat{j} + (1-3)\hat{k} = -4\hat{i} - 3\hat{j} - 2\hat{k}$
$\vec{AC} = (2-1)\hat{i} + (1-2)\hat{j} + (3-3)\hat{k} = 1\hat{i} - 1\hat{j} + 0\hat{k}$
$\vec{AD} = (-1-1)\hat{i} + (2-2)\hat{j} + (x-3)\hat{k} = -2\hat{i} + 0\hat{j} + (x-3)\hat{k}$
The scalar triple product is the determinant:
$|\vec{AB} \vec{AC} \vec{AD}| = \begin{vmatrix} -4 & -3 & -2 \\ 1 & -1 & 0 \\ -2 & 0 & x-3 \end{vmatrix}$
$= -4(-1(x-3) - 0) - (-3)(1(x-3) - 0) + (-2)(0 - 2)$
$= -4(-x+3) + 3(x-3) + 4$
$= 4x - 12 + 3x - 9 + 4 = 7x - 17$
Given volume $V = \frac{11}{6}$,so $\frac{1}{6} |7x - 17| = \frac{11}{6} \implies |7x - 17| = 11$.
Case $1$: $7x - 17 = 11 \implies 7x = 28 \implies x = 4$.
Case $2$: $7x - 17 = -11 \implies 7x = 6 \implies x = \frac{6}{7}$.
Comparing with the given options,$x = 4$ is the correct value.

(D) Given that $\overrightarrow{r} \cdot \overrightarrow{a} = 0$,this implies that $\overrightarrow{r}$ is perpendicular to $\overrightarrow{a}$.
From $|\overrightarrow{r} \times \overrightarrow{b}| = |\overrightarrow{r}| |\overrightarrow{b}|$,we know that $\sin \theta = 1$,where $\theta$ is the angle between $\overrightarrow{r}$ and $\overrightarrow{b}$. Thus,$\overrightarrow{r}$ is perpendicular to $\overrightarrow{b}$.
Similarly,from $|\overrightarrow{r} \times \overrightarrow{c}| = |\overrightarrow{r}| |\overrightarrow{c}|$,we conclude that $\overrightarrow{r}$ is perpendicular to $\overrightarrow{c}$.
Since $\overrightarrow{r}$ is a non-zero vector perpendicular to $\overrightarrow{a}, \overrightarrow{b},$ and $\overrightarrow{c}$,these three vectors must lie in a plane perpendicular to $\overrightarrow{r}$.
Therefore,$\overrightarrow{a}, \overrightarrow{b},$ and $\overrightarrow{c}$ are coplanar.
For any three coplanar vectors,their scalar triple product is zero.
Thus,$[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}] = 0$.

(C) The scalar triple product is defined as $[\vec{a} \vec{b} \vec{c}] = \vec{a} \cdot (\vec{b} \times \vec{c})$.
Since $\vec{a}$ is perpendicular to both $\vec{b}$ and $\vec{c}$,$\vec{a}$ must be parallel to the vector $\vec{b} \times \vec{c}$.
Thus,the angle between $\vec{a}$ and $\vec{b} \times \vec{c}$ is $0$ or $\pi$.
Assuming $\vec{a}$ is in the direction of $\vec{b} \times \vec{c}$,we have:
$[\vec{a} \vec{b} \vec{c}] = |\vec{a}| |\vec{b} \times \vec{c}| \cos(0) = |\vec{a}| |\vec{b}| |\vec{c}| \sin\left(\frac{5 \pi}{6}\right)$.
Substituting the given values:
$[\vec{a} \vec{b} \vec{c}] = 5 \times 3 \times 4 \times \sin\left(\frac{5 \pi}{6}\right)$.
Since $\sin\left(\frac{5 \pi}{6}\right) = \sin\left(\pi - \frac{\pi}{6}\right) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$.
$[\vec{a} \vec{b} \vec{c}] = 60 \times \frac{1}{2} = 30$.

(A) The scalar triple product $[\vec{a} \vec{b} \vec{c}]$ 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:
$[\vec{a} \vec{b} \vec{c}] = 1 \cdot \begin{vmatrix} 1 & 1 \\ x & 1+x \end{vmatrix} - 0 + 0$
$[\vec{a} \vec{b} \vec{c}] = (1+x) - x = 1$
Since the result is $1$,which is a constant,the value of $[\vec{a} \vec{b} \vec{c}]$ depends on neither $x$ nor $y$.

(D) The volume of a tetrahedron with vertices $A(x_1, y_1, z_1)$,$B(x_2, y_2, z_2)$,$C(x_3, y_3, z_3)$,and $D(x_4, y_4, z_4)$ is given by the formula:
$V = \frac{1}{6} |(\vec{AB}) \cdot (\vec{AC} \times \vec{AD})| = \frac{1}{6} \left| \det \begin{bmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ x_3-x_1 & y_3-y_1 & z_3-z_1 \\ x_4-x_1 & y_4-y_1 & z_4-z_1 \end{bmatrix} \right|$
Given vertices are $A(1, -6, 10)$,$B(-1, -3, 7)$,$C(5, -1, \lambda)$,and $D(7, -4, 7)$.
Calculating the vectors:
$\vec{AB} = (-1-1, -3-(-6), 7-10) = (-2, 3, -3)$
$\vec{AC} = (5-1, -1-(-6), \lambda-10) = (4, 5, \lambda-10)$
$\vec{AD} = (7-1, -4-(-6), 7-10) = (6, 2, -3)$
The volume is $11$,so:
$11 = \frac{1}{6} \left| \det \begin{bmatrix} -2 & 3 & -3 \\ 4 & 5 & \lambda-10 \\ 6 & 2 & -3 \end{bmatrix} \right|$
$66 = | -2(-15 - 2(\lambda-10)) - 3(-12 - 6(\lambda-10)) - 3(8 - 30) |$
$66 = | -2(-15 - 2\lambda + 20) - 3(-12 - 6\lambda + 60) - 3(-22) |$
$66 = | -2(5 - 2\lambda) - 3(48 - 6\lambda) + 66 |$
$66 = | -10 + 4\lambda - 144 + 18\lambda + 66 |$
$66 = | 22\lambda - 88 |$
Dividing by $22$: $3 = | \lambda - 4 |$
This gives two cases:
$1) \lambda - 4 = 3 \Rightarrow \lambda = 7$
$2) \lambda - 4 = -3 \Rightarrow \lambda = 1$
Since the options provided include $7$,we select $\lambda = 7$.

(A) The volume $V$ of a parallelepiped formed by vectors $\vec{u}, \vec{v}, \vec{w}$ is given by the absolute value of their scalar triple product: $V = |\vec{u} \cdot (\vec{v} \times \vec{w})|$.
The vectors are $\vec{u} = (1, a, 1)$,$\vec{v} = (0, 1, a)$,and $\vec{w} = (a, 0, 1)$.
The scalar triple product is given by the determinant:
$V(a) = \left|\begin{array}{ccc} 1 & a & 1 \\ 0 & 1 & a \\ a & 0 & 1 \end{array}\right|$
Expanding along the first row:
$V(a) = 1(1 - 0) - a(0 - a^2) + 1(0 - a) = 1 + a^3 - a = a^3 - a + 1$.
To find the minimum volume,we find the derivative $V'(a)$ and set it to $0$:
$V'(a) = 3a^2 - 1 = 0$.
$3a^2 = 1 \Rightarrow a^2 = \frac{1}{3} \Rightarrow a = \pm \frac{1}{\sqrt{3}}$.
Using the second derivative test: $V''(a) = 6a$.
For $a = \frac{1}{\sqrt{3}}$,$V''(a) = 6(\frac{1}{\sqrt{3}}) > 0$,which indicates a local minimum.
Therefore,the volume is minimum at $a = \frac{1}{\sqrt{3}}$.

(C) The volume of a tetrahedron with edges $\overline{u}, \overline{v}, \overline{w}$ is given by $V = \frac{1}{6} |[\overline{u} \overline{v} \overline{w}]|$.
Given edges are $\overline{a}+\overline{b}, \overline{b}+\overline{c}, \overline{c}+\overline{a}$.
So,$24 = \frac{1}{6} |(\overline{a}+\overline{b}) \cdot ((\overline{b}+\overline{c}) \times (\overline{c}+\overline{a}))|$.
$144 = |(\overline{a}+\overline{b}) \cdot (\overline{b} \times \overline{c} + \overline{b} \times \overline{a} + \overline{c} \times \overline{c} + \overline{c} \times \overline{a})|$.
Since $\overline{c} \times \overline{c} = 0$,we have $144 = |(\overline{a}+\overline{b}) \cdot (\overline{b} \times \overline{c} + \overline{b} \times \overline{a} + \overline{c} \times \overline{a})|$.
Expanding the scalar triple product: $144 = |[\overline{a} \overline{b} \overline{c}] + [\overline{a} \overline{b} \overline{a}] + [\overline{a} \overline{c} \overline{a}] + [\overline{b} \overline{b} \overline{c}] + [\overline{b} \overline{b} \overline{a}] + [\overline{b} \overline{c} \overline{a}]|$.
Terms like $[\overline{a} \overline{b} \overline{a}]$ are $0$ because two vectors are identical.
Thus,$144 = |[\overline{a} \overline{b} \overline{c}] + [\overline{b} \overline{c} \overline{a}]|$.
Since $[\overline{a} \overline{b} \overline{c}] = [\overline{b} \overline{c} \overline{a}]$,we get $144 = 2 |[\overline{a} \overline{b} \overline{c}]|$.
Therefore,$|[\overline{a} \overline{b} \overline{c}]| = 72$,which is the volume of the parallelepiped.

(D) The scalar triple product is defined as $[\vec{u} \vec{v} \vec{w}] = \vec{u} \cdot (\vec{v} \times \vec{w})$.
First,calculate the cross product $\vec{v} \times \vec{w}$:
$\vec{v} \times \vec{w} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 2 & -1 \\ 1 & 0 & 3 \end{vmatrix} = \hat{i}(6 - 0) - \hat{j}(6 - (-1)) + \hat{k}(0 - 2) = 6\hat{i} - 7\hat{j} - 2\hat{k}$.
The magnitude of this vector is $|\vec{v} \times \vec{w}| = \sqrt{6^2 + (-7)^2 + (-2)^2} = \sqrt{36 + 49 + 4} = \sqrt{89}$.
Since $\vec{u}$ is a unit vector $(|\vec{u}| = 1)$,the scalar triple product is $\vec{u} \cdot (\vec{v} \times \vec{w}) = |\vec{u}| |\vec{v} \times \vec{w}| \cos \theta$,where $\theta$ is the angle between $\vec{u}$ and $(\vec{v} \times \vec{w})$.
The maximum value occurs when $\cos \theta = 1$,which gives the maximum value as $|\vec{v} \times \vec{w}| = \sqrt{89}$.

(D) We are given the expression $\bar{a} \cdot [(\bar{b} \times \bar{c}) \times (\bar{a} + \bar{b} + \bar{c})]$.
Using the distributive property of the cross product,we expand the term inside the square brackets:
$(\bar{b} \times \bar{c}) \times (\bar{a} + \bar{b} + \bar{c}) = (\bar{b} \times \bar{c}) \times \bar{a} + (\bar{b} \times \bar{c}) \times \bar{b} + (\bar{b} \times \bar{c}) \times \bar{c}$.
Now,take the dot product with $\bar{a}$:
$\bar{a} \cdot [(\bar{b} \times \bar{c}) \times \bar{a} + (\bar{b} \times \bar{c}) \times \bar{b} + (\bar{b} \times \bar{c}) \times \bar{c}]$
$= \bar{a} \cdot ((\bar{b} \times \bar{c}) \times \bar{a}) + \bar{a} \cdot ((\bar{b} \times \bar{c}) \times \bar{b}) + \bar{a} \cdot ((\bar{b} \times \bar{c}) \times \bar{c})$.
Using the scalar triple product property $\bar{x} \cdot (\bar{y} \times \bar{z}) = [\bar{x} \bar{y} \bar{z}]$,we note that if any two vectors in the triple product are the same,the value is $0$.
$1$. $\bar{a} \cdot ((\bar{b} \times \bar{c}) \times \bar{a}) = [\bar{a} (\bar{b} \times \bar{c}) \bar{a}] = 0$ (since $\bar{a}$ is repeated).
$2$. $\bar{a} \cdot ((\bar{b} \times \bar{c}) \times \bar{b}) = [\bar{a} (\bar{b} \times \bar{c}) \bar{b}] = 0$ (since $\bar{b}$ is repeated).
$3$. $\bar{a} \cdot ((\bar{b} \times \bar{c}) \times \bar{c}) = [\bar{a} (\bar{b} \times \bar{c}) \bar{c}] = 0$ (since $\bar{c}$ is repeated).
Thus,the entire expression equals $0+0+0 = 0$.