MHT CET 2020 Mathematics Question Paper with Answer and Solution

(A) Given $y=\tan ^{-1}\left(\frac{\sin 2 x}{1+\cos 2 x}\right)$.
Using trigonometric identities $\sin 2x = 2 \sin x \cos x$ and $1 + \cos 2x = 2 \cos^2 x$,we get:
$y = \tan^{-1}\left(\frac{2 \sin x \cos x}{2 \cos^2 x}\right)$
$y = \tan^{-1}(\tan x)$
$y = x$
Therefore,$\frac{dy}{dx} = \frac{d}{dx}(x) = 1$.

(B) Given the equation $3 \bar{a}-\bar{b}+2 \bar{c}-4 \bar{d}=\overline{0}$.
Rearranging the terms,we get $3 \bar{a}+2 \bar{c}=\bar{b}+4 \bar{d}$.
Dividing both sides by $5$,we have $\frac{3 \bar{a}+2 \bar{c}}{5}=\frac{\bar{b}+4 \bar{d}}{5}$.
Let $\bar{r} = \frac{3 \bar{a}+2 \bar{c}}{3+2} = \frac{\bar{b}+4 \bar{d}}{1+4}$.
This vector $\bar{r}$ represents a point that lies on the line segment $AC$ (dividing it in ratio $2:3$) and also on the line segment $BD$ (dividing it in ratio $4:1$).
Thus,$\bar{r}$ is the position vector of the point of intersection of $AC$ and $BD$.

(B) Let $D$ be the midpoint of $BC$. The median through $A$ is the vector $\vec{AD}$.
Since $D$ is the midpoint of $BC$,the position vector of $D$ relative to $A$ is given by $\vec{AD} = \frac{1}{2}(\vec{AB} + \vec{AC})$.
Given $\vec{AB} = 3 \hat{i} + 5 \hat{j} + 4 \hat{k}$ and $\vec{AC} = 5 \hat{i} - 5 \hat{j} + 2 \hat{k}$.
Substituting these values:
$\vec{AD} = \frac{1}{2}((3 \hat{i} + 5 \hat{j} + 4 \hat{k}) + (5 \hat{i} - 5 \hat{j} + 2 \hat{k}))$
$\vec{AD} = \frac{1}{2}(8 \hat{i} + 0 \hat{j} + 6 \hat{k})$
$\vec{AD} = 4 \hat{i} + 3 \hat{k}$.
The length of the median is the magnitude of vector $\vec{AD}$:
$|\vec{AD}| = \sqrt{4^2 + 0^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$ units.

(B) Let the position vector of $H$ be the origin $\vec{0}$.
Then the position vectors of the vertices $A, B, C$ are $\vec{a}, \vec{b}, \vec{c}$ respectively.
The given equation is $a \vec{AH} + b \vec{BH} + c \vec{CH} = \vec{0}$.
Since $H$ is the origin,$\vec{AH} = \vec{0} - \vec{a} = -\vec{a}$,$\vec{BH} = -\vec{b}$,and $\vec{CH} = -\vec{c}$.
Substituting these into the equation,we get $a(-\vec{a}) + b(-\vec{b}) + c(-\vec{c}) = \vec{0}$,which implies $a\vec{a} + b\vec{b} + c\vec{c} = \vec{0}$.
However,the standard definition of the incentre $I$ with position vector $\vec{i}$ is $\vec{i} = \frac{a\vec{A} + b\vec{B} + c\vec{C}}{a+b+c}$.
If $H$ is the incentre,then $a(\vec{A}-\vec{H}) + b(\vec{B}-\vec{H}) + c(\vec{C}-\vec{H}) = \vec{0}$.
This simplifies to $(a+b+c)\vec{H} = a\vec{A} + b\vec{B} + c\vec{C}$,which is the definition of the incentre.
Therefore,$H$ is the incentre of $\triangle ABC$.

(B) Since the vectors $\bar{a}$,$\bar{b}$,and $\bar{c}$ are coplanar,their scalar triple product must be zero,i.e.,$\bar{a} \cdot (\bar{b} \times \bar{c}) = 0$.
This is equivalent to the determinant of the matrix formed by their components being zero:
$\begin{vmatrix} 1 & -2 & 1 \\ 2 & -5 & p \\ 5 & -9 & 4 \end{vmatrix} = 0$.
Expanding the determinant along the first row:
$1((-5)(4) - (-9)(p)) - (-2)((2)(4) - (5)(p)) + 1((2)(-9) - (5)(-5)) = 0$.
$1(-20 + 9p) + 2(8 - 5p) + 1(-18 + 25) = 0$.
$-20 + 9p + 16 - 10p + 7 = 0$.
$-p + 3 = 0$.
Therefore,$p = 3$.

(A) Let $\vec{a} = \hat{i} + 2\hat{j} + x\hat{k}$ and $\vec{b} = y\hat{i} + 6\hat{j} + 4\hat{k}$ be two collinear vectors.
Since they are collinear,there exists a scalar $m$ such that $\vec{a} = m\vec{b}$.
Substituting the components,we get: $\hat{i} + 2\hat{j} + x\hat{k} = m(y\hat{i} + 6\hat{j} + 4\hat{k})$.
Equating the coefficients of $\hat{i}, \hat{j},$ and $\hat{k}$ on both sides:
$1 = my$
$2 = 6m \Rightarrow m = \frac{2}{6} = \frac{1}{3}$
$x = 4m$
Substituting $m = \frac{1}{3}$ into the equations for $x$ and $y$:
$x = 4 \times \frac{1}{3} = \frac{4}{3}$
$1 = \frac{1}{3}y \Rightarrow y = 3$
Thus,the values are $x = \frac{4}{3}$ and $y = 3$.

(B) Since the vectors $\bar{a}$,$\bar{b}$,and $\bar{c}$ are coplanar,their scalar triple product must be zero: $\bar{a} \cdot (\bar{b} \times \bar{c}) = 0$.
This is equivalent to the determinant of the components being zero:
$\begin{vmatrix} 1 & 1 & 1 \\ 1 & -1 & 2 \\ x & x-1 & -1 \end{vmatrix} = 0$.
Expanding the determinant along the first row:
$1((-1)(-1) - (2)(x-1)) - 1((1)(-1) - (2)(x)) + 1((1)(x-1) - (-1)(x)) = 0$.
$1(1 - 2x + 2) - 1(-1 - 2x) + 1(x - 1 + x) = 0$.
$(3 - 2x) + (1 + 2x) + (2x - 1) = 0$.
$3 - 2x + 1 + 2x + 2x - 1 = 0$.
$2x + 3 = 0$.
$x = \frac{-3}{2}$.

(C) Since the given vectors are coplanar,their scalar triple product must be zero.
$\therefore \begin{vmatrix} a & a & c \\ 1 & 0 & 1 \\ c & c & b \end{vmatrix} = 0$
Expanding the determinant along the second row:
$-1(ab - c^2) + 0(ab - c^2) - 1(ac - ac) = 0$
Wait,expanding along the first row:
$a(0 - c) - a(b - c) + c(c - 0) = 0$
$-ac - ab + ac + c^2 = 0$
$-ab + c^2 = 0$
$c^2 = ab$
This implies that $a, c, b$ are in $G$.$P$.

(A) The equation of the line passing through points $A(3, 5, -1)$ and $B(6, 3, -2)$ is given by the formula $\frac{x-x_1}{x_2-x_1} = \frac{y-y_1}{y_2-y_1} = \frac{z-z_1}{z_2-z_1}$.
Substituting the coordinates,we get $\frac{x-3}{6-3} = \frac{y-5}{3-5} = \frac{z-(-1)}{-2-(-1)}$,which simplifies to $\frac{x-3}{3} = \frac{y-5}{-2} = \frac{z+1}{-1} = r$.
Any point $P$ on this line can be represented as $(3r+3, -2r+5, -r-1)$.
Given that the $y$-coordinate of point $P$ is $2$,we set $-2r+5 = 2$.
Solving for $r$,we get $-2r = -3$,so $r = \frac{3}{2}$.
Now,substitute $r = \frac{3}{2}$ into the $z$-coordinate expression: $z = -r-1 = -\frac{3}{2} - 1 = -\frac{5}{2}$.

(C) Since the given vectors $\overline{a}, \overline{b},$ and $\overline{c}$ are coplanar,their scalar triple product must be zero,i.e.,$[\overline{a} \overline{b} \overline{c}] = 0$.
This is equivalent to the determinant of the matrix formed by their components being zero:
$\begin{vmatrix} 2 & -1 & 1 \\ 1 & 2 & -3 \\ 3 & \lambda & 5 \end{vmatrix} = 0$
Expanding the determinant along the first row:
$2(2 \times 5 - (-3) \times \lambda) - (-1)(1 \times 5 - (-3) \times 3) + 1(1 \times \lambda - 2 \times 3) = 0$
$2(10 + 3\lambda) + 1(5 + 9) + 1(\lambda - 6) = 0$
$20 + 6\lambda + 14 + \lambda - 6 = 0$
$7\lambda + 28 = 0$
$7\lambda = -28 \Rightarrow \lambda = -4$
Now,we check which equation has $x = -4$ as a root:
$(A)$ $(-4)^2 + 2(-4) = 16 - 8 = 8 \neq 6$
$(B)$ $(-4)^2 + 2(-4) = 16 - 8 = 8 \neq 4$
$(C)$ $(-4)^2 + 3(-4) = 16 - 12 = 4$. This is correct.
$(D)$ $(-4)^2 + 3(-4) = 16 - 12 = 4 \neq 6$
Thus,$\lambda = -4$ is the root of the equation $x^2 + 3x = 4$.

(B) Let the vertices be $A$,$B$,and $C$ with position vectors $\vec{a} = \hat{i}+\hat{j}+\hat{k}$,$\vec{b} = 5\hat{i}+3\hat{j}-3\hat{k}$,and $\vec{c} = 2\hat{i}+5\hat{j}+9\hat{k}$.
The side lengths are the magnitudes of the vectors $\vec{AB}$,$\vec{BC}$,and $\vec{AC}$.
$\vec{AB} = \vec{b} - \vec{a} = (5-1)\hat{i} + (3-1)\hat{j} + (-3-1)\hat{k} = 4\hat{i} + 2\hat{j} - 4\hat{k}$.
$|\vec{AB}| = \sqrt{4^2 + 2^2 + (-4)^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6$.
$\vec{BC} = \vec{c} - \vec{b} = (2-5)\hat{i} + (5-3)\hat{j} + (9-(-3))\hat{k} = -3\hat{i} + 2\hat{j} + 12\hat{k}$.
$|\vec{BC}| = \sqrt{(-3)^2 + 2^2 + 12^2} = \sqrt{9 + 4 + 144} = \sqrt{157}$.
$\vec{AC} = \vec{c} - \vec{a} = (2-1)\hat{i} + (5-1)\hat{j} + (9-1)\hat{k} = 1\hat{i} + 4\hat{j} + 8\hat{k}$.
$|\vec{AC}| = \sqrt{1^2 + 4^2 + 8^2} = \sqrt{1 + 16 + 64} = \sqrt{81} = 9$.
The perimeter of the triangle is $|\vec{AB}| + |\vec{BC}| + |\vec{AC}| = 6 + \sqrt{157} + 9 = 15 + \sqrt{157}$ units.

(B) Let $\vec{a}, \vec{b}, \vec{c}, \vec{d}, \vec{m}, \vec{n}$ be the position vectors of $A, B, C, D, M, N$ respectively.
Since $M$ and $N$ are the midpoints of $AB$ and $CD$ respectively,we have:
$\vec{m} = \frac{\vec{a} + \vec{b}}{2} \implies \vec{a} + \vec{b} = 2\vec{m}$
$\vec{n} = \frac{\vec{c} + \vec{d}}{2} \implies \vec{c} + \vec{d} = 2\vec{n}$
We are given the equation: $\vec{AD} + \vec{BC} = t \vec{MN}$
Expressing the vectors in terms of position vectors:
$(\vec{d} - \vec{a}) + (\vec{c} - \vec{b}) = t(\vec{n} - \vec{m})$
Rearranging the terms:
$(\vec{d} + \vec{c}) - (\vec{a} + \vec{b}) = t(\vec{n} - \vec{m})$
Substituting the expressions for the sums of position vectors:
$2\vec{n} - 2\vec{m} = t(\vec{n} - \vec{m})$
$2(\vec{n} - \vec{m}) = t(\vec{n} - \vec{m})$
Comparing both sides,we get $t = 2$.

(A) Let $\vec{a}, \vec{b}, \vec{c}$ be the position vectors of vertices $A, B, C$ with respect to the origin $O$.
Then,$\overline{OA} = \vec{a}, \overline{OB} = \vec{b}, \overline{OC} = \vec{c}$.
The centroid $G$ of triangle $ABC$ is given by the position vector $\overline{OG} = \frac{\vec{a} + \vec{b} + \vec{c}}{3}$.
This implies $\vec{a} + \vec{b} + \vec{c} = 3 \overline{OG}$.
Now,we evaluate the expression $\overline{OA} + \overline{OB} + \overline{OC} + \overline{OG}$:
$\overline{OA} + \overline{OB} + \overline{OC} + \overline{OG} = (\vec{a} + \vec{b} + \vec{c}) + \overline{OG}$.
Substituting the value of $(\vec{a} + \vec{b} + \vec{c})$:
$= 3 \overline{OG} + \overline{OG} = 4 \overline{OG}$.

(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 $b$.

(B) Since $\bar{a}$ and $\bar{b}$ are non-collinear vectors,two vectors $\bar{p} = m_1\bar{a} + n_1\bar{b}$ and $\bar{q} = m_2\bar{a} + n_2\bar{b}$ are collinear if and only if their components are proportional,i.e.,$\frac{m_1}{m_2} = \frac{n_1}{n_2} = k$.
Given $\bar{p} = (2x + 1)\bar{a} - 1\bar{b}$ and $\bar{q} = (x - 2)\bar{a} + 1\bar{b}$.
For collinearity,we have:
$\frac{2x + 1}{x - 2} = \frac{-1}{1}$
$2x + 1 = -(x - 2)$
$2x + 1 = -x + 2$
$3x = 1$
$x = \frac{1}{3}$

(A) The equation of the line passing through $A(2, 4, 5)$ and $B(1, 2, 3)$ is given by:
$\frac{x-2}{1-2} = \frac{y-4}{2-4} = \frac{z-5}{3-5} = k$
$\frac{x-2}{-1} = \frac{y-4}{-2} = \frac{z-5}{-2} = k$
The coordinates of any point $P$ on this line are given by $(x, y, z) = (-k+2, -2k+4, -2k+5)$.
Given that the $z$-coordinate of point $P$ is $3$,we have:
$-2k + 5 = 3$
$-2k = -2$
$k = 1$
Substituting $k = 1$ into the expression for the $y$-coordinate:
$y = -2(1) + 4 = 2$.
Thus,the $y$-coordinate of point $P$ is $2$.

(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.

(A) The position vector of the centroid $G$ of a triangle with vertices having position vectors $\overline{a}$,$\overline{b}$,and $\overline{c}$ is given by the formula: $\overline{g} = \frac{\overline{a} + \overline{b} + \overline{c}}{3}$.
Substituting the given vectors:
$\overline{g} = \frac{(2 \hat{\imath} + 3 \hat{\jmath} + \hat{k}) + (4 \hat{\imath} + 5 \hat{\jmath} + 3 \hat{k}) + (6 \hat{\imath} + \hat{\jmath} + 5 \hat{k})}{3}$
Summing the components:
$\overline{g} = \frac{(2+4+6) \hat{\imath} + (3+5+1) \hat{\jmath} + (1+3+5) \hat{k}}{3}$
$\overline{g} = \frac{12 \hat{\imath} + 9 \hat{\jmath} + 9 \hat{k}}{3}$
$\overline{g} = 4 \hat{\imath} + 3 \hat{\jmath} + 3 \hat{k}$.

(D) Let the position vectors of vertices $A, B, C$ be $\vec{a} = 4 \hat{\imath} + 7 \hat{\jmath} + 8 \hat{k}$,$\vec{b} = 2 \hat{\imath} + 3 \hat{\jmath} + 4 \hat{k}$,and $\vec{c} = 2 \hat{\imath} + 5 \hat{\jmath} + 7 \hat{k}$.
According to the Angle Bisector Theorem,the bisector of $\angle A$ divides the opposite side $BC$ in the ratio of the sides $AB$ and $AC$.
Let $D$ be the point where the bisector of $\angle A$ meets $BC$. Then $D$ divides $BC$ in the ratio $AB : AC$.
First,calculate the lengths of sides $AB$ and $AC$:
$\vec{AB} = \vec{b} - \vec{a} = (2-4)\hat{\imath} + (3-7)\hat{\jmath} + (4-8)\hat{k} = -2\hat{\imath} - 4\hat{\jmath} - 4\hat{k}$.
$AB = |\vec{AB}| = \sqrt{(-2)^2 + (-4)^2 + (-4)^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$.
$\vec{AC} = \vec{c} - \vec{a} = (2-4)\hat{\imath} + (5-7)\hat{\jmath} + (7-8)\hat{k} = -2\hat{\imath} - 2\hat{\jmath} - 1\hat{k}$.
$AC = |\vec{AC}| = \sqrt{(-2)^2 + (-2)^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.
The ratio $AB : AC = 6 : 3 = 2 : 1$.
Using the section formula,the position vector of $D$ is given by:
$\vec{d} = \frac{AB \cdot \vec{c} + AC \cdot \vec{b}}{AB + AC} = \frac{6(2\hat{\imath} + 5\hat{\jmath} + 7\hat{k}) + 3(2\hat{\imath} + 3\hat{\jmath} + 4\hat{k})}{6 + 3}$
$\vec{d} = \frac{(12\hat{\imath} + 30\hat{\jmath} + 42\hat{k}) + (6\hat{\imath} + 9\hat{\jmath} + 12\hat{k})}{9} = \frac{18\hat{\imath} + 39\hat{\jmath} + 54\hat{k}}{9} = 2\hat{\imath} + \frac{13}{3}\hat{\jmath} + 6\hat{k}$.
Wait,re-evaluating the ratio division: $D$ divides $BC$ in ratio $c:b$ (where $c=AB, b=AC$).
$\vec{d} = \frac{AC \cdot \vec{b} + AB \cdot \vec{c}}{AC + AB} = \frac{3(2\hat{\imath} + 3\hat{\jmath} + 4\hat{k}) + 6(2\hat{\imath} + 5\hat{\jmath} + 7\hat{k})}{3 + 6} = \frac{6\hat{\imath} + 9\hat{\jmath} + 12\hat{k} + 12\hat{\imath} + 30\hat{\jmath} + 42\hat{k}}{9} = \frac{18\hat{\imath} + 39\hat{\jmath} + 54\hat{k}}{9} = 2\hat{\imath} + \frac{13}{3}\hat{\jmath} + 6\hat{k}$.
Checking options: $\frac{1}{3}(6\hat{\imath} + 13\hat{\jmath} + 18\hat{k}) = 2\hat{\imath} + \frac{13}{3}\hat{\jmath} + 6\hat{k}$. This matches option $D$.

(D) Given position vectors are $\bar{a} = \hat{i} + 3\hat{j}$,$\bar{b} = 2\hat{i} + 5\hat{j}$,and $\bar{c} = 4\hat{i} + 2\hat{j}$.
Given the relation $\bar{c} = t_{1}\bar{a} + t_{2}\bar{b}$,we substitute the vectors:
$4\hat{i} + 2\hat{j} = t_{1}(\hat{i} + 3\hat{j}) + t_{2}(2\hat{i} + 5\hat{j})$
$4\hat{i} + 2\hat{j} = (t_{1} + 2t_{2})\hat{i} + (3t_{1} + 5t_{2})\hat{j}$
Comparing the coefficients of $\hat{i}$ and $\hat{j}$,we get the system of linear equations:
$t_{1} + 2t_{2} = 4$ --- $(1)$
$3t_{1} + 5t_{2} = 2$ --- $(2)$
Multiply equation $(1)$ by $3$: $3t_{1} + 6t_{2} = 12$ --- $(3)$
Subtract equation $(2)$ from $(3)$: $(3t_{1} + 6t_{2}) - (3t_{1} + 5t_{2}) = 12 - 2$
$t_{2} = 10$
Substitute $t_{2} = 10$ into equation $(1)$: $t_{1} + 2(10) = 4$
$t_{1} + 20 = 4 \implies t_{1} = -16$
Therefore,$t_{1}t_{2} = (-16)(10) = -160$.

(C) Given the equation $l \bar{a} + m \bar{b} + n \bar{c} = \overline{0}$.
Substituting the vectors: $l(\hat{\imath} + \hat{\jmath} + \hat{k}) + m(2\hat{\imath} - 2\hat{\jmath} + 2\hat{k}) + n(2\hat{\imath} + 3\hat{\jmath} + 2\hat{k}) = 0\hat{\imath} + 0\hat{\jmath} + 0\hat{k}$.
Equating the components to zero:
$l + 2m + 2n = 0$ $(i)$
$l - 2m + 3n = 0$ (ii)
$l + 2m + 2n = 0$ (iii)
From $(i)$ and (iii),we see they are identical. Subtracting (ii) from $(i)$:
$(l + 2m + 2n) - (l - 2m + 3n) = 0$
$4m - n = 0 \implies n = 4m$.
Substitute $n = 4m$ into $(i)$:
$l + 2m + 2(4m) = 0 \implies l + 10m = 0 \implies l = -10m$.
Let $m = -1$,then $l = 10$ and $n = -4$.
Thus,the values are $(10, -1, -4)$.

(A) First,calculate the cross product $\bar{a} \times \bar{b}$:
$\bar{a} \times \bar{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -1 \\ -1 & 2 & -4 \end{vmatrix} = \hat{i}(-12 - (-2)) - \hat{j}(-8 - 1) + \hat{k}(4 - (-3)) = -10\hat{i} + 9\hat{j} + 7\hat{k}$
Next,calculate the cross product $\bar{a} \times \bar{c}$:
$\bar{a} \times \bar{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -1 \\ 1 & 1 & 1 \end{vmatrix} = \hat{i}(3 - (-1)) - \hat{j}(2 - (-1)) + \hat{k}(2 - 3) = 4\hat{i} - 3\hat{j} - \hat{k}$
Now,compute the dot product of the two resulting vectors:
$(\bar{a} \times \bar{b}) \cdot (\bar{a} \times \bar{c}) = (-10\hat{i} + 9\hat{j} + 7\hat{k}) \cdot (4\hat{i} - 3\hat{j} - \hat{k})$
$= (-10)(4) + (9)(-3) + (7)(-1)$
$= -40 - 27 - 7 = -74$

(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) Let $\vec{a}$ and $\vec{b}$ be the vectors along the lines whose direction ratios are $2, 3, 1$ and $1, 2, 1$ respectively.
$\vec{a} = 2\hat{i} + 3\hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + \hat{k}$.
$A$ vector perpendicular to both $\vec{a}$ and $\vec{b}$ is given by their cross product $\vec{a} \times \vec{b}$.
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 1 \\ 1 & 2 & 1 \end{vmatrix} = \hat{i}(3-2) - \hat{j}(2-1) + \hat{k}(4-3) = \hat{i} - \hat{j} + \hat{k}$.
Thus,the direction ratios are $1, -1, 1$.
Comparing this with the given options,the correct choice is $A$.

(C) Let the given vectors be $\vec{a} = \hat{\imath} - \hat{\jmath} - 6\hat{k}$,$\vec{b} = \hat{\imath} - 3\hat{\jmath} + 4\hat{k}$,and $\vec{c} = 2\hat{\imath} - 5\hat{\jmath} + m\hat{k}$.
Since the 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:
$\begin{vmatrix} 1 & -1 & -6 \\ 1 & -3 & 4 \\ 2 & -5 & m \end{vmatrix} = 0$
Expanding along the first row:
$1((-3)(m) - (4)(-5)) - (-1)((1)(m) - (4)(2)) + (-6)((1)(-5) - (-3)(2)) = 0$
$1(-3m + 20) + 1(m - 8) - 6(-5 + 6) = 0$
$-3m + 20 + m - 8 - 6(1) = 0$
$-2m + 12 - 6 = 0$
$-2m + 6 = 0$
$2m = 6$
$m = 3$

(A) The volume of a parallelepiped with coterminous edges $\vec{u}, \vec{v}, \vec{w}$ is given by the scalar triple product $[\vec{u} \quad \vec{v} \quad \vec{w}] = \vec{u} \cdot (\vec{v} \times \vec{w})$.
Given $[\bar{a} \quad \bar{b} \quad \bar{c}] = 4$.
We need to find the volume of the parallelepiped with edges $\bar{a}+2 \bar{b}$,$\bar{b}+2 \bar{c}$,and $\bar{c}+2 \bar{a}$.
Volume $= [(\bar{a}+2 \bar{b}) \quad (\bar{b}+2 \bar{c}) \quad (\bar{c}+2 \bar{a})]$.
$= (\bar{a}+2 \bar{b}) \cdot [(\bar{b}+2 \bar{c}) \times (\bar{c}+2 \bar{a})]$.
$= (\bar{a}+2 \bar{b}) \cdot [(\bar{b} \times \bar{c}) + 2(\bar{b} \times \bar{a}) + 2(\bar{c} \times \bar{c}) + 4(\bar{c} \times \bar{a})]$.
Since $\bar{c} \times \bar{c} = 0$,this simplifies to:
$= (\bar{a}+2 \bar{b}) \cdot [(\bar{b} \times \bar{c}) + 2(\bar{b} \times \bar{a}) + 4(\bar{c} \times \bar{a})]$.
$= \bar{a} \cdot (\bar{b} \times \bar{c}) + 2\bar{a} \cdot (\bar{b} \times \bar{a}) + 4\bar{a} \cdot (\bar{c} \times \bar{a}) + 2\bar{b} \cdot (\bar{b} \times \bar{c}) + 4\bar{b} \cdot (\bar{b} \times \bar{a}) + 8\bar{b} \cdot (\bar{c} \times \bar{a})$.
Using the property that the scalar triple product is zero if any two vectors are the same:
$= [\bar{a} \quad \bar{b} \quad \bar{c}] + 0 + 0 + 0 + 0 + 8[\bar{b} \quad \bar{c} \quad \bar{a}]$.
Since $[\bar{b} \quad \bar{c} \quad \bar{a}] = [\bar{a} \quad \bar{b} \quad \bar{c}] = 4$,
Volume $= 4 + 8(4) = 4 + 32 = 36 \text{ units}^3$.

(C) Given vectors $\bar{a}$ and $\bar{b}$ are unit vectors as $|\bar{a}| = \sqrt{\frac{9+1}{10}} = 1$ and $|\bar{b}| = \sqrt{\frac{4+9+36}{49}} = 1$.
First,calculate the dot product $\bar{a} \cdot \bar{b} = \frac{1}{7\sqrt{10}}(3 \times 2 + 0 \times 3 + 1 \times (-6)) = \frac{6-6}{7\sqrt{10}} = 0$.
Now,simplify the expression $(2 \bar{a}-\bar{b}) \cdot[(\bar{a} \times \bar{b}) \times(\bar{a}+2 \bar{b})]$.
Using the vector triple product identity $\bar{u} \times (\bar{v} \times \bar{w}) = (\bar{u} \cdot \bar{w}) \bar{v} - (\bar{u} \cdot \bar{v}) \bar{w}$,we have:
$(\bar{a} \times \bar{b}) \times(\bar{a}+2 \bar{b}) = -[(\bar{a}+2 \bar{b}) \times (\bar{a} \times \bar{b})] = -[(\bar{a}+2 \bar{b}) \cdot \bar{b}] \bar{a} + [(\bar{a}+2 \bar{b}) \cdot \bar{a}] \bar{b}$.
Since $\bar{a} \cdot \bar{b} = 0$,$|\bar{a}|=1$,and $|\bar{b}|=1$:
$(\bar{a}+2 \bar{b}) \cdot \bar{b} = \bar{a} \cdot \bar{b} + 2|\bar{b}|^2 = 0 + 2(1) = 2$.
$(\bar{a}+2 \bar{b}) \cdot \bar{a} = |\bar{a}|^2 + 2(\bar{b} \cdot \bar{a}) = 1 + 0 = 1$.
So,the expression becomes $-(2 \bar{a}-\bar{b}) \cdot [-2 \bar{a} + \bar{b}] = (2 \bar{a}-\bar{b}) \cdot (2 \bar{a}-\bar{b}) = |2 \bar{a}-\bar{b}|^2$.
$|2 \bar{a}-\bar{b}|^2 = 4|\bar{a}|^2 + |\bar{b}|^2 - 4(\bar{a} \cdot \bar{b}) = 4(1) + 1 - 4(0) = 5$.

(C) First,calculate the cross product $\bar{u} \times \bar{v}$:
$\bar{u} \times \bar{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 1 \\ 3 & 0 & 1 \end{vmatrix} = \hat{i}(-2-0) - \hat{j}(1-3) + \hat{k}(0 - (-6)) = -2\hat{i} + 2\hat{j} + 6\hat{k}$.
Next,define the vectors for the edges:
$\bar{a} = \bar{u} \times \bar{v} = -2\hat{i} + 2\hat{j} + 6\hat{k}$
$\bar{b} = \bar{u} + \bar{w} = (\hat{i} - 2\hat{j} + \hat{k}) + (\hat{j} - \hat{k}) = \hat{i} - \hat{j}$
$\bar{c} = \bar{v} + \bar{w} = (3\hat{i} + \hat{k}) + (\hat{j} - \hat{k}) = 3\hat{i} + \hat{j}$
The volume of the parallelepiped is given by the scalar triple product $|\bar{a} \cdot (\bar{b} \times \bar{c})|$,which is equivalent to the absolute value of the determinant of the matrix formed by these vectors:
Volume $= \left| \begin{vmatrix} -2 & 2 & 6 \\ 1 & -1 & 0 \\ 3 & 1 & 0 \end{vmatrix} \right|$
Expanding along the third column:
Volume $= |6(1 - (-3))| = |6(4)| = 24$ cubic units.

(A) The volume of a tetrahedron with vertices $A, B, C, D$ is given by the formula $V = \frac{1}{6} |[\vec{AB} \vec{AC} \vec{AD}]|$.
First,we find the vectors:
$\vec{AB} = (3 - (-1))\hat{i} + (-2 - 2)\hat{j} + (1 - 3)\hat{k} = 4\hat{i} - 4\hat{j} - 2\hat{k}$
$\vec{AC} = (2 - (-1))\hat{i} + (1 - 2)\hat{j} + (3 - 3)\hat{k} = 3\hat{i} - 1\hat{j} + 0\hat{k}$
$\vec{AD} = (-1 - (-1))\hat{i} + (-2 - 2)\hat{j} + (4 - 3)\hat{k} = 0\hat{i} - 4\hat{j} + 1\hat{k}$
Now,calculate the scalar triple product $[\vec{AB} \vec{AC} \vec{AD}]$ using the determinant:
$|\vec{AB} \vec{AC} \vec{AD}| = \begin{vmatrix} 4 & -4 & -2 \\ 3 & -1 & 0 \\ 0 & -4 & 1 \end{vmatrix}$
$= 4((-1)(1) - (0)(-4)) - (-4)((3)(1) - (0)(0)) + (-2)((3)(-4) - (-1)(0))$
$= 4(-1) + 4(3) - 2(-12) = -4 + 12 + 24 = 32$
Therefore,the volume $V = \frac{1}{6} \times 32 = \frac{32}{6} = \frac{16}{3}$ cu. units.

(C) The scalar triple product is defined as $[\bar{a} \quad \bar{b} \quad \bar{c}] = \bar{a} \cdot (\bar{b} \times \bar{c})$.
We need to evaluate the expression $[\bar{a}+2\bar{b} \quad \bar{b}+2\bar{c} \quad \bar{c}+2\bar{a}]$.
Using the properties of the scalar triple product,we can write this as:
$[\bar{a}+2\bar{b} \quad \bar{b}+2\bar{c} \quad \bar{c}+2\bar{a}] = (\bar{a}+2\bar{b}) \cdot [(\bar{b}+2\bar{c}) \times (\bar{c}+2\bar{a})]$.
Expanding the cross product:
$(\bar{b}+2\bar{c}) \times (\bar{c}+2\bar{a}) = (\bar{b} \times \bar{c}) + 2(\bar{b} \times \bar{a}) + 2(\bar{c} \times \bar{c}) + 4(\bar{c} \times \bar{a})$.
Since $\bar{c} \times \bar{c} = 0$,this simplifies to $(\bar{b} \times \bar{c}) + 2(\bar{b} \times \bar{a}) + 4(\bar{c} \times \bar{a})$.
Now,take the dot product with $(\bar{a}+2\bar{b})$:
$(\bar{a}+2\bar{b}) \cdot [(\bar{b} \times \bar{c}) + 2(\bar{b} \times \bar{a}) + 4(\bar{c} \times \bar{a})]$
$= \bar{a} \cdot (\bar{b} \times \bar{c}) + 2\bar{a} \cdot (\bar{b} \times \bar{a}) + 4\bar{a} \cdot (\bar{c} \times \bar{a}) + 2\bar{b} \cdot (\bar{b} \times \bar{c}) + 4\bar{b} \cdot (\bar{b} \times \bar{a}) + 8\bar{b} \cdot (\bar{c} \times \bar{a})$.
Using the property that the scalar triple product is zero if any two vectors are identical:
$= [\bar{a} \bar{b} \bar{c}] + 0 + 0 + 0 + 0 + 8[\bar{b} \bar{c} \bar{a}]$
$= [\bar{a} \bar{b} \bar{c}] + 8[\bar{a} \bar{b} \bar{c}] = 9[\bar{a} \bar{b} \bar{c}]$.
Therefore,$\frac{[\bar{a}+2\bar{b} \quad \bar{b}+2\bar{c} \quad \bar{c}+2\bar{a}]}{[\bar{a} \bar{b} \bar{c}]} = \frac{9[\bar{a} \bar{b} \bar{c}]}{[\bar{a} \bar{b} \bar{c}]} = 9$.

(C) The volume of a parallelepiped with coterminus edges $\vec{u}, \vec{v}, \vec{w}$ is given by the scalar triple product $[\vec{u} \vec{v} \vec{w}] = \vec{u} \cdot (\vec{v} \times \vec{w})$.
Given $\vec{u} = 2\bar{a}+\bar{b}$,$\vec{v} = 2\bar{b}+\vec{c}$,and $\vec{w} = 2\bar{c}+\bar{a}$.
Volume $= (2\bar{a}+\bar{b}) \cdot [(2\bar{b}+\bar{c}) \times (2\bar{c}+\bar{a})]$.
Expanding the cross product: $(2\bar{b}+\bar{c}) \times (2\bar{c}+\bar{a}) = 4(\bar{b} \times \bar{c}) + 2(\bar{b} \times \bar{a}) + 2(\bar{c} \times \bar{c}) + (\bar{c} \times \bar{a}) = 4(\bar{b} \times \bar{c}) + 2(\bar{b} \times \bar{a}) + (\bar{c} \times \bar{a})$.
Now,taking the dot product with $(2\bar{a}+\bar{b})$:
Volume $= (2\bar{a}+\bar{b}) \cdot [4(\bar{b} \times \bar{c}) + 2(\bar{b} \times \bar{a}) + (\bar{c} \times \bar{a})]$.
$= 8[\bar{a} \bar{b} \bar{c}] + 4[\bar{a} \bar{b} \bar{a}] + 2[\bar{a} \bar{c} \bar{a}] + 4[\bar{b} \bar{b} \bar{c}] + 2[\bar{b} \bar{b} \bar{a}] + [\bar{b} \bar{c} \bar{a}]$.
Since the scalar triple product is zero if any two vectors are identical,we have $[\bar{a} \bar{b} \bar{a}] = 0, [\bar{a} \bar{c} \bar{a}] = 0, [\bar{b} \bar{b} \bar{c}] = 0, [\bar{b} \bar{b} \bar{a}] = 0$.
Volume $= 8[\bar{a} \bar{b} \bar{c}] + [\bar{b} \bar{c} \bar{a}] = 8[\bar{a} \bar{b} \bar{c}] + [\bar{a} \bar{b} \bar{c}] = 9[\bar{a} \bar{b} \bar{c}]$.
Given $[\bar{a} \bar{b} \bar{c}] = 3$,Volume $= 9 \times 3 = 27$ cubic units.

(B) The volume of a tetrahedron with vertices $A, B, C, D$ is given by $V = \frac{1}{6} |(\vec{AB}) \cdot (\vec{AC} \times \vec{AD})|$.
First,we find the vectors:
$\vec{AB} = (3 - (-1))\hat{i} + (-2 - 2)\hat{j} + (1 - 3)\hat{k} = 4\hat{i} - 4\hat{j} - 2\hat{k}$
$\vec{AC} = (2 - (-1))\hat{i} + (1 - 2)\hat{j} + (3 - 3)\hat{k} = 3\hat{i} - 1\hat{j} + 0\hat{k}$
$\vec{AD} = (-1 - (-1))\hat{i} + (-2 - 2)\hat{j} + (4 - 3)\hat{k} = 0\hat{i} - 4\hat{j} + 1\hat{k}$
Now,calculate the scalar triple product:
$V = \frac{1}{6} \left| \begin{vmatrix} 4 & -4 & -2 \\ 3 & -1 & 0 \\ 0 & -4 & 1 \end{vmatrix} \right|$
$V = \frac{1}{6} |4(-1 - 0) - (-4)(3 - 0) + (-2)(-12 - 0)|$
$V = \frac{1}{6} |4(-1) + 4(3) - 2(-12)|$
$V = \frac{1}{6} |-4 + 12 + 24| = \frac{1}{6} |32| = \frac{32}{6} = \frac{16}{3} \text{ cu. units}$.

(C) We know that the scalar triple product $[\bar{a}+\bar{b} \quad \bar{b}+\bar{c} \quad \bar{c}+\bar{a}]$ can be expanded as follows:
$[\bar{a}+\bar{b} \quad \bar{b}+\bar{c} \quad \bar{c}+\bar{a}] = (\bar{a}+\bar{b}) \cdot [(\bar{b}+\bar{c}) \times (\bar{c}+\bar{a})]$
$= (\bar{a}+\bar{b}) \cdot [(\bar{b} \times \bar{c}) + (\bar{b} \times \bar{a}) + (\bar{c} \times \bar{c}) + (\bar{c} \times \bar{a})]$
Since $\bar{c} \times \bar{c} = 0$,we have:
$= (\bar{a}+\bar{b}) \cdot [(\bar{b} \times \bar{c}) + (\bar{b} \times \bar{a}) + (\bar{c} \times \bar{a})]$
$= \bar{a} \cdot (\bar{b} \times \bar{c}) + \bar{a} \cdot (\bar{b} \times \bar{a}) + \bar{a} \cdot (\bar{c} \times \bar{a}) + \bar{b} \cdot (\bar{b} \times \bar{c}) + \bar{b} \cdot (\bar{b} \times \bar{a}) + \bar{b} \cdot (\bar{c} \times \bar{a})$
Using the property that the scalar triple product is zero if any two vectors are identical,we get:
$= [\bar{a} \bar{b} \bar{c}] + 0 + 0 + 0 + 0 + [\bar{b} \bar{c} \bar{a}]$
Since $[\bar{a} \bar{b} \bar{c}] = [\bar{b} \bar{c} \bar{a}]$,we have:
$= [\bar{a} \bar{b} \bar{c}] + [\bar{a} \bar{b} \bar{c}] = 2[\bar{a} \bar{b} \bar{c}]$
Therefore,the given expression is $\frac{2[\bar{a} \bar{b} \bar{c}]}{[\bar{a} \bar{b} \bar{c}]} = 2$.

(C) For three vectors to be coplanar,their scalar triple product must be zero.
Let the vectors be $\vec{a} = \hat{\imath}+\hat{\jmath}+\hat{k}$,$\vec{b} = \hat{\imath}-\hat{\jmath}+\hat{k}$,and $\vec{c} = 2\hat{\imath}+3\hat{\jmath}+m\hat{k}$.
The condition for coplanarity is $\vec{a} \cdot (\vec{b} \times \vec{c}) = 0$,which is equivalent to the determinant of the components being zero:
$\begin{vmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 2 & 3 & m \end{vmatrix} = 0$
Expanding the determinant along the first row:
$1((-1)(m) - (1)(3)) - 1((1)(m) - (1)(2)) + 1((1)(3) - (-1)(2)) = 0$
$1(-m - 3) - 1(m - 2) + 1(3 + 2) = 0$
$-m - 3 - m + 2 + 5 = 0$
$-2m + 4 = 0$
$2m = 4$
$m = 2$

(A) Given that the volume of the parallelepiped formed by vectors $\bar{a}, \bar{b}, \bar{c}$ is $12$. Thus,the scalar triple product is $[\bar{a} \bar{b} \bar{c}] = 12$.
The volume of a tetrahedron with coterminous edges $\bar{u}, \bar{v}, \bar{w}$ is given by $\frac{1}{6} |[\bar{u} \bar{v} \bar{w}]|$.
Here,the edges are $\bar{a}+\bar{b}, \bar{b}+\bar{c}, \bar{c}+\bar{a}$.
We calculate the scalar triple product:
$[\bar{a}+\bar{b} \quad \bar{b}+\bar{c} \quad \bar{c}+\bar{a}] = (\bar{a}+\bar{b}) \cdot ((\bar{b}+\bar{c}) \times (\bar{c}+\bar{a}))$
$= (\bar{a}+\bar{b}) \cdot (\bar{b} \times \bar{c} + \bar{b} \times \bar{a} + \bar{c} \times \bar{c} + \bar{c} \times \bar{a})$
Since $\bar{c} \times \bar{c} = 0$,this simplifies to:
$= (\bar{a}+\bar{b}) \cdot (\bar{b} \times \bar{c} + \bar{b} \times \bar{a} + \bar{c} \times \bar{a})$
$= \bar{a} \cdot (\bar{b} \times \bar{c}) + \bar{a} \cdot (\bar{b} \times \bar{a}) + \bar{a} \cdot (\bar{c} \times \bar{a}) + \bar{b} \cdot (\bar{b} \times \bar{c}) + \bar{b} \cdot (\bar{b} \times \bar{a}) + \bar{b} \cdot (\bar{c} \times \bar{a})$
$= [\bar{a} \bar{b} \bar{c}] + 0 + 0 + 0 + 0 + [\bar{b} \bar{c} \bar{a}]$
$= [\bar{a} \bar{b} \bar{c}] + [\bar{a} \bar{b} \bar{c}] = 2[\bar{a} \bar{b} \bar{c}] = 2(12) = 24$.
Therefore,the volume of the tetrahedron is $\frac{1}{6} \times 24 = 4 \text{ (units)}^3$.

(B) The scalar triple product is defined as $[\bar{b}, \bar{a} \times \bar{b}, \bar{a}] = \bar{b} \cdot ((\bar{a} \times \bar{b}) \times \bar{a})$.
Using the property of the vector triple product,$(\bar{a} \times \bar{b}) \times \bar{a} = -\bar{a} \times (\bar{a} \times \bar{b})$.
By the vector triple product formula $\bar{u} \times (\bar{v} \times \bar{w}) = (\bar{u} \cdot \bar{w})\bar{v} - (\bar{u} \cdot \bar{v})\bar{w}$,we have $\bar{a} \times (\bar{a} \times \bar{b}) = (\bar{a} \cdot \bar{b})\bar{a} - (\bar{a} \cdot \bar{a})\bar{b}$.
Thus,$[\bar{b}, \bar{a} \times \bar{b}, \bar{a}] = \bar{b} \cdot ((\bar{a} \cdot \bar{b})\bar{a} - (\bar{a} \cdot \bar{a})\bar{b}) = (\bar{a} \cdot \bar{b})(\bar{b} \cdot \bar{a}) - (\bar{a} \cdot \bar{a})(\bar{b} \cdot \bar{b}) = (\bar{a} \cdot \bar{b})^2 - |\bar{a}|^2 |\bar{b}|^2$.
Using Lagrange's identity,$|\bar{a} \times \bar{b}|^2 = |\bar{a}|^2 |\bar{b}|^2 - (\bar{a} \cdot \bar{b})^2$.
Therefore,$[\bar{b}, \bar{a} \times \bar{b}, \bar{a}] = -|\bar{a} \times \bar{b}|^2$.

(C) The volume of a parallelepiped with coterminus edges $\bar{a}, \bar{b}, \bar{c}$ is given by the scalar triple product $[\bar{a} \bar{b} \bar{c}] = |\bar{a} \cdot (\bar{b} \times \bar{c})| = 7$.
We need to find the volume of a parallelepiped with edges $\bar{a}+\bar{b}, \bar{b}+\bar{c}, \bar{c}+\bar{a}$.
This volume is given by the scalar triple product $[\bar{a}+\bar{b} \quad \bar{b}+\bar{c} \quad \bar{c}+\bar{a}]$.
Using the property of the scalar triple product,we have:
$[\bar{a}+\bar{b} \quad \bar{b}+\bar{c} \quad \bar{c}+\bar{a}] = (\bar{a}+\bar{b}) \cdot ((\bar{b}+\bar{c}) \times (\bar{c}+\bar{a}))$.
Expanding the cross product:
$(\bar{b}+\bar{c}) \times (\bar{c}+\bar{a}) = \bar{b} \times \bar{c} + \bar{b} \times \bar{a} + \bar{c} \times \bar{c} + \bar{c} \times \bar{a} = \bar{b} \times \bar{c} + \bar{b} \times \bar{a} + \bar{c} \times \bar{a}$ (since $\bar{c} \times \bar{c} = 0$).
Now,taking the dot product with $(\bar{a}+\bar{b})$:
$(\bar{a}+\bar{b}) \cdot (\bar{b} \times \bar{c} + \bar{b} \times \bar{a} + \bar{c} \times \bar{a}) = \bar{a} \cdot (\bar{b} \times \bar{c}) + \bar{a} \cdot (\bar{b} \times \bar{a}) + \bar{a} \cdot (\bar{c} \times \bar{a}) + \bar{b} \cdot (\bar{b} \times \bar{c}) + \bar{b} \cdot (\bar{b} \times \bar{a}) + \bar{b} \cdot (\bar{c} \times \bar{a})$.
Since the scalar triple product is zero if any two vectors are the same,we have:
$= [\bar{a} \bar{b} \bar{c}] + 0 + 0 + 0 + 0 + [\bar{b} \bar{c} \bar{a}] = [\bar{a} \bar{b} \bar{c}] + [\bar{a} \bar{b} \bar{c}] = 2[\bar{a} \bar{b} \bar{c}]$.
Given $[\bar{a} \bar{b} \bar{c}] = 7$,the volume is $2 \times 7 = 14$ cubic units.

(C) Given that $\bar{p}=\frac{\bar{b} \times \bar{c}}{[\bar{a} \bar{b} \bar{c}]}, \bar{q}=\frac{\bar{c} \times \bar{a}}{[\bar{a} \bar{b} \bar{c}]}, \bar{r}=\frac{\bar{a} \times \bar{b}}{[\bar{a} \bar{b} \bar{c}]}$.
We know that the scalar triple product $[\bar{a} \bar{b} \bar{c}] = \bar{a} \cdot (\bar{b} \times \bar{c})$.
Now,calculate each term:
$\bar{a} \cdot \bar{p} = \bar{a} \cdot \left( \frac{\bar{b} \times \bar{c}}{[\bar{a} \bar{b} \bar{c}]} \right) = \frac{\bar{a} \cdot (\bar{b} \times \bar{c})}{[\bar{a} \bar{b} \bar{c}]} = \frac{[\bar{a} \bar{b} \bar{c}]}{[\bar{a} \bar{b} \bar{c}]} = 1$.
Similarly,$\bar{b} \cdot \bar{q} = \bar{b} \cdot \left( \frac{\bar{c} \times \bar{a}}{[\bar{a} \bar{b} \bar{c}]} \right) = \frac{\bar{b} \cdot (\bar{c} \times \bar{a})}{[\bar{a} \bar{b} \bar{c}]} = \frac{[\bar{b} \bar{c} \bar{a}]}{[\bar{a} \bar{b} \bar{c}]} = \frac{[\bar{a} \bar{b} \bar{c}]}{[\bar{a} \bar{b} \bar{c}]} = 1$.
And $\bar{c} \cdot \bar{r} = \bar{c} \cdot \left( \frac{\bar{a} \times \bar{b}}{[\bar{a} \bar{b} \bar{c}]} \right) = \frac{\bar{c} \cdot (\bar{a} \times \bar{b})}{[\bar{a} \bar{b} \bar{c}]} = \frac{[\bar{c} \bar{a} \bar{b}]}{[\bar{a} \bar{b} \bar{c}]} = \frac{[\bar{a} \bar{b} \bar{c}]}{[\bar{a} \bar{b} \bar{c}]} = 1$.
Therefore,$\bar{a} \cdot \bar{p} + \bar{b} \cdot \bar{q} + \bar{c} \cdot \bar{r} = 1 + 1 + 1 = 3$.

(D) We expand the dot product: $(\bar{a} + \bar{b} + \bar{c}) \cdot (\bar{a} \times \bar{b} + \bar{b} \times \bar{c} + \bar{c} \times \bar{a})$.
Using the distributive property of the dot product,we get:
$= \bar{a} \cdot (\bar{a} \times \bar{b}) + \bar{a} \cdot (\bar{b} \times \bar{c}) + \bar{a} \cdot (\bar{c} \times \bar{a}) + \bar{b} \cdot (\bar{a} \times \bar{b}) + \bar{b} \cdot (\bar{b} \times \bar{c}) + \bar{b} \cdot (\bar{c} \times \bar{a}) + \bar{c} \cdot (\bar{a} \times \bar{b}) + \bar{c} \cdot (\bar{b} \times \bar{c}) + \bar{c} \cdot (\bar{c} \times \bar{a})$.
Since the scalar triple product $[\bar{x} \bar{y} \bar{z}] = 0$ if any two vectors are identical,terms like $\bar{a} \cdot (\bar{a} \times \bar{b}) = 0$,$\bar{b} \cdot (\bar{b} \times \bar{c}) = 0$,and $\bar{c} \cdot (\bar{c} \times \bar{a}) = 0$.
This leaves us with:
$= [\bar{a} \bar{b} \bar{c}] + [\bar{b} \bar{c} \bar{a}] + [\bar{c} \bar{a} \bar{b}]$.
Since the scalar triple product is invariant under cyclic permutation,$[\bar{a} \bar{b} \bar{c}] = [\bar{b} \bar{c} \bar{a}] = [\bar{c} \bar{a} \bar{b}]$.
Therefore,the expression equals $3[\bar{a} \bar{b} \bar{c}]$.
Comparing this with $k[\bar{a} \bar{b} \bar{c}]$,we find $k = 3$.