MHT CET 2024 Mathematics Question Paper with Answer and Solution

(A) Let $\bar{v} = 2\bar{a} - \bar{b} + 3\bar{c}$.
Substituting the given vectors:
$\bar{v} = 2(\hat{i} + \hat{j} + \hat{k}) - (4\hat{i} - 2\hat{j} + 3\hat{k}) + 3(\hat{i} - 2\hat{j} + \hat{k})$
$\bar{v} = (2 - 4 + 3)\hat{i} + (2 + 2 - 6)\hat{j} + (2 - 3 + 3)\hat{k}$
$\bar{v} = \hat{i} - 2\hat{j} + 2\hat{k}$
The magnitude of $\bar{v}$ is $|\bar{v}| = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.
The unit vector in the direction of $\bar{v}$ is $\hat{v} = \frac{\bar{v}}{|\bar{v}|} = \frac{1}{3}(\hat{i} - 2\hat{j} + 2\hat{k})$.
The required vector of magnitude $6$ is $6 \times \hat{v} = 6 \times \frac{1}{3}(\hat{i} - 2\hat{j} + 2\hat{k}) = 2(\hat{i} - 2\hat{j} + 2\hat{k}) = 2\hat{i} - 4\hat{j} + 4\hat{k}$.

(A) The position vector is $\overline{OP} = \hat{a} \cos t + \hat{b} \sin t$.
The length $M = |\overline{OP}|$ is given by:
$M^2 = |\hat{a} \cos t + \hat{b} \sin t|^2 = |\hat{a}|^2 \cos^2 t + |\hat{b}|^2 \sin^2 t + 2(\hat{a} \cdot \hat{b}) \sin t \cos t$.
Since $\hat{a}$ and $\hat{b}$ are unit vectors,$|\hat{a}| = |\hat{b}| = 1$.
$M^2 = \cos^2 t + \sin^2 t + (\hat{a} \cdot \hat{b}) \sin 2t = 1 + (\hat{a} \cdot \hat{b}) \sin 2t$.
For $M$ to be maximum,$\sin 2t$ must be $1$,so $2t = \frac{\pi}{2}$,which means $t = \frac{\pi}{4}$.
Then $M = \sqrt{1 + \hat{a} \cdot \hat{b}}$.
The unit vector $\hat{u}$ along $\overline{OP}$ is $\frac{\overline{OP}}{|\overline{OP}|}$.
At $t = \frac{\pi}{4}$,$\overline{OP} = \hat{a} \cos(\frac{\pi}{4}) + \hat{b} \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}(\hat{a} + \hat{b})$.
Thus,$\hat{u} = \frac{\frac{1}{\sqrt{2}}(\hat{a} + \hat{b})}{|\frac{1}{\sqrt{2}}(\hat{a} + \hat{b})|} = \frac{\hat{a} + \hat{b}}{|\hat{a} + \hat{b}|}$.

(A) Given $\bar{x}=\frac{\bar{b} \times \bar{c}}{[\bar{a} \bar{b} \bar{c}]}, \bar{y}=\frac{\bar{c} \times \bar{a}}{[\bar{a} \bar{b} \bar{c}]}, \bar{z}=\frac{\bar{a} \times \bar{b}}{[\bar{a} \bar{b} \bar{c}]}$.
We need to evaluate $S = \bar{x} \cdot(\bar{a}+\bar{b})+\bar{y} \cdot(\bar{b}+\bar{c})+\bar{z} \cdot(\bar{c}+\bar{a})$.
Substituting the values of $\bar{x}, \bar{y}, \bar{z}$:
$S = \frac{(\bar{b} \times \bar{c}) \cdot (\bar{a}+\bar{b})}{[\bar{a} \bar{b} \bar{c}]} + \frac{(\bar{c} \times \bar{a}) \cdot (\bar{b}+\bar{c})}{[\bar{a} \bar{b} \bar{c}]} + \frac{(\bar{a} \times \bar{b}) \cdot (\bar{c}+\bar{a})}{[\bar{a} \bar{b} \bar{c}]}$.
Using the property of scalar triple product $[\bar{a} \bar{b} \bar{c}] = (\bar{a} \times \bar{b}) \cdot \bar{c}$ and the fact that the scalar triple product is zero if any two vectors are identical:
$S = \frac{[\bar{b} \bar{c} \bar{a}] + [\bar{b} \bar{c} \bar{b}] + [\bar{c} \bar{a} \bar{b}] + [\bar{c} \bar{a} \bar{c}] + [\bar{a} \bar{b} \bar{c}] + [\bar{a} \bar{b} \bar{a}]}{[\bar{a} \bar{b} \bar{c}]}$.
Since $[\bar{b} \bar{c} \bar{b}] = 0, [\bar{c} \bar{a} \bar{c}] = 0, [\bar{a} \bar{b} \bar{a}] = 0$ and $[\bar{b} \bar{c} \bar{a}] = [\bar{c} \bar{a} \bar{b}] = [\bar{a} \bar{b} \bar{c}]$,we get:
$S = \frac{[\bar{a} \bar{b} \bar{c}] + 0 + [\bar{a} \bar{b} \bar{c}] + 0 + [\bar{a} \bar{b} \bar{c}] + 0}{[\bar{a} \bar{b} \bar{c}]} = \frac{3[\bar{a} \bar{b} \bar{c}]}{[\bar{a} \bar{b} \bar{c}]} = 3$.

(D) Given that $\overline{a}+2\overline{b}$ is collinear with $\overline{c}$,we have $\overline{a}+2\overline{b} = n\overline{c}$ for some non-zero scalar $n$. $(i)$
Similarly,$\overline{b}+3\overline{c}$ is collinear with $\overline{a}$,so $\overline{b}+3\overline{c} = m\overline{a}$ for some non-zero scalar $m$. (ii)
From (ii),we have $\overline{b} = m\overline{a} - 3\overline{c}$.
Substitute this into $(i)$: $\overline{a} + 2(m\overline{a} - 3\overline{c}) = n\overline{c}$.
This simplifies to $(1+2m)\overline{a} = (n+6)\overline{c}$.
Since $\overline{a}$ and $\overline{c}$ are not collinear,the coefficients must be zero: $1+2m = 0 \Rightarrow m = -1/2$ and $n+6 = 0 \Rightarrow n = -6$.
Now,consider the expression $\overline{a}+2\overline{b}+6\overline{c}$.
From $(i)$,$\overline{a}+2\overline{b} = n\overline{c} = -6\overline{c}$.
Therefore,$\overline{a}+2\overline{b}+6\overline{c} = -6\overline{c} + 6\overline{c} = \overline{0}$.

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

(A) Let the position vectors of points $P, Q$ and $R$ be $\vec{p} = \hat{i}-2 \hat{j}+3 \hat{k}$,$\vec{q} = -2 \hat{i}+3 \hat{j}+2 \hat{k}$ and $\vec{r} = -8 \hat{i}+13 \hat{j}$.
First,we find the vectors $\vec{PQ}$ and $\vec{QR}$:
$\vec{PQ} = \vec{q} - \vec{p} = (-2 \hat{i}+3 \hat{j}+2 \hat{k}) - (\hat{i}-2 \hat{j}+3 \hat{k}) = -3 \hat{i}+5 \hat{j}-\hat{k}$.
$\vec{QR} = \vec{r} - \vec{q} = (-8 \hat{i}+13 \hat{j}) - (-2 \hat{i}+3 \hat{j}+2 \hat{k}) = -6 \hat{i}+10 \hat{j}-2 \hat{k}$.
We observe that $\vec{QR} = 2(-3 \hat{i}+5 \hat{j}-\hat{k}) = 2 \vec{PQ}$.
Since $\vec{QR}$ is a scalar multiple of $\vec{PQ}$,the vectors $\vec{PQ}$ and $\vec{QR}$ are parallel.
Because they share a common point $Q$,the points $P, Q$ and $R$ are collinear.
Since $\vec{QR} = 2 \vec{PQ}$,the point $Q$ lies between $P$ and $R$.

(D) Let $AD$ be the median of $\triangle ABC$ through vertex $A$.
Since $D$ is the midpoint of $BC$,the vector $\overline{AD}$ is given by the formula $\overline{AD} = \frac{\overline{AB} + \overline{AC}}{2}$.
Substituting the given vectors:
$\overline{AD} = \frac{(3 \hat{i} + 4 \hat{k}) + (5 \hat{i} - 2 \hat{j} + 4 \hat{k})}{2}$
$\overline{AD} = \frac{(3+5) \hat{i} + (-2) \hat{j} + (4+4) \hat{k}}{2}$
$\overline{AD} = \frac{8 \hat{i} - 2 \hat{j} + 8 \hat{k}}{2}$
$\overline{AD} = 4 \hat{i} - \hat{j} + 4 \hat{k}$
Now,the length of the median is the magnitude of vector $\overline{AD}$:
$|\overline{AD}| = \sqrt{4^2 + (-1)^2 + 4^2}$
$|\overline{AD}| = \sqrt{16 + 1 + 16}$
$|\overline{AD}| = \sqrt{33} \text{ units}$.

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

(C) Given vectors are $\bar{a}=2 \hat{i}+2 \hat{j}+3 \hat{k}$,$\bar{b}=-\hat{i}+2 \hat{j}+\hat{k}$ and $\bar{c}=3 \hat{i}+\hat{j}$.
First,we find the vector $\bar{a}+\lambda \bar{b}$:
$\bar{a}+\lambda \bar{b} = (2 \hat{i}+2 \hat{j}+3 \hat{k}) + \lambda(-\hat{i}+2 \hat{j}+\hat{k})$
$= (2-\lambda) \hat{i} + (2+2\lambda) \hat{j} + (3+\lambda) \hat{k}$.
Since $\bar{a}+\lambda \bar{b}$ is perpendicular to $\bar{c}$,their dot product must be zero:
$(\bar{a}+\lambda \bar{b}) \cdot \bar{c} = 0$
$((2-\lambda) \hat{i} + (2+2\lambda) \hat{j} + (3+\lambda) \hat{k}) \cdot (3 \hat{i} + 1 \hat{j} + 0 \hat{k}) = 0$
$(2-\lambda)(3) + (2+2\lambda)(1) + (3+\lambda)(0) = 0$
$6 - 3\lambda + 2 + 2\lambda = 0$
$8 - \lambda = 0$
$\lambda = 8$.

(A) First,calculate the vector $\vec{v} = 3 \bar{a} + \bar{b} - 2 \bar{c}$.
$\vec{v} = 3(2 \hat{i} - \hat{j} + \hat{k}) + (\hat{i} + \hat{j} - 2 \hat{k}) - 2(4 \hat{i} - 2 \hat{j} + \hat{k})$
$\vec{v} = (6 \hat{i} - 3 \hat{j} + 3 \hat{k}) + (\hat{i} + \hat{j} - 2 \hat{k}) - (8 \hat{i} - 4 \hat{j} + 2 \hat{k})$
$\vec{v} = (6 + 1 - 8) \hat{i} + (-3 + 1 + 4) \hat{j} + (3 - 2 - 2) \hat{k}$
$\vec{v} = -\hat{i} + 2 \hat{j} - \hat{k}$
Now,find the magnitude of $\vec{v}$:
$|\vec{v}| = \sqrt{(-1)^2 + 2^2 + (-1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6}$
The unit vector in the direction of $\vec{v}$ is given by $\frac{\vec{v}}{|\vec{v}|} = \frac{-\hat{i} + 2 \hat{j} - \hat{k}}{\sqrt{6}} = \frac{1}{\sqrt{6}}(-\hat{i} + 2 \hat{j} - \hat{k})$.

(C) Let $\vec{a} = Cx \hat{i} - 6 \hat{j} - 3 \hat{k}$ and $\vec{b} = x \hat{i} + 2 \hat{j} + 2Cx \hat{k}$.
Since the angle between $\vec{a}$ and $\vec{b}$ is obtuse,their dot product must be negative,i.e.,$\vec{a} \cdot \vec{b} < 0$.
Calculating the dot product: $(Cx)(x) + (-6)(2) + (-3)(2Cx) < 0$.
$Cx^2 - 12 - 6Cx < 0$.
$Cx^2 - 6Cx - 12 < 0$.
For this quadratic expression in $x$ to be negative for all real $x$,the coefficient of $x^2$ must be negative $(C < 0)$ and the discriminant $D$ must be negative $(D < 0)$.
$D = (-6C)^2 - 4(C)(-12) = 36C^2 + 48C < 0$.
Dividing by $12$: $3C^2 + 4C < 0$.
$C(3C + 4) < 0$.
This inequality holds when $-\frac{4}{3} < C < 0$.
Since we also require $C < 0$,the intersection is $(-\frac{4}{3}, 0)$.

(B) $f(x) = \bar{a} \cdot (\bar{b} \times \bar{c}) = \begin{vmatrix} x & -2 & 3 \\ -2 & x & -1 \\ 7 & -2 & x \end{vmatrix}$
Expanding the determinant:
$f(x) = x(x^2 - 2) + 2(-2x + 7) + 3(4 - 7x)$
$f(x) = x^3 - 2x - 4x + 14 + 12 - 21x$
$f(x) = x^3 - 27x + 26$
To find local minima,we find $f'(x) = 0$:
$f'(x) = 3x^2 - 27 = 0$
$3(x^2 - 9) = 0 \Rightarrow x = \pm 3$
Using the second derivative test:
$f''(x) = 6x$
$f''(3) = 18 > 0$,so $x_0 = 3$ is the point of local minima.
At $x = 3$:
$\bar{a} = 3 \hat{i} - 2 \hat{j} + 3 \hat{k}$
$\bar{b} = -2 \hat{i} + 3 \hat{j} - \hat{k}$
$\bar{a} \cdot \bar{b} = (3)(-2) + (-2)(3) + (3)(-1)$
$\bar{a} \cdot \bar{b} = -6 - 6 - 3 = -15$

(B) Given vectors are $\bar{a}=(2 \hat{i}+2 \hat{j}+3 \hat{k})$,$\bar{b}=(-\hat{i}+2 \hat{j}+\hat{k})$,and $\bar{c}=(3 \hat{i}+\hat{j})$.
First,calculate the vector $(\bar{a}+\lambda \bar{b})$:
$(\bar{a}+\lambda \bar{b}) = (2 \hat{i}+2 \hat{j}+3 \hat{k}) + \lambda(-\hat{i}+2 \hat{j}+\hat{k})$
$= (2-\lambda) \hat{i} + (2+2\lambda) \hat{j} + (3+\lambda) \hat{k}$.
Since $(\bar{a}+\lambda \bar{b})$ is perpendicular to $\bar{c}$,their dot product must be zero:
$(\bar{a}+\lambda \bar{b}) \cdot \bar{c} = 0$
$((2-\lambda) \hat{i} + (2+2\lambda) \hat{j} + (3+\lambda) \hat{k}) \cdot (3 \hat{i}+1 \hat{j}+0 \hat{k}) = 0$
$(2-\lambda)(3) + (2+2\lambda)(1) + (3+\lambda)(0) = 0$
$6 - 3\lambda + 2 + 2\lambda = 0$
$8 - \lambda = 0$
$\lambda = 8$.

(D) Given vectors are $\bar{a}=\hat{i}-\hat{j}+2\hat{k}$,$\bar{b}=2\hat{i}+4\hat{j}+\hat{k}$ and $\bar{c}=m\hat{i}+\hat{j}+n\hat{k}$.
Since the vectors are mutually perpendicular,$\bar{a} \cdot \bar{c} = 0$ and $\bar{b} \cdot \bar{c} = 0$.
For $\bar{a} \cdot \bar{c} = 0$:
$(1)(m) + (-1)(1) + (2)(n) = 0$
$m - 1 + 2n = 0$
$m + 2n = 1$ ... $(i)$
For $\bar{b} \cdot \bar{c} = 0$:
$(2)(m) + (4)(1) + (1)(n) = 0$
$2m + n = -4$ ... $(ii)$
Multiplying equation $(ii)$ by $2$,we get $4m + 2n = -8$ ... $(iii)$.
Subtracting $(i)$ from $(iii)$:
$(4m + 2n) - (m + 2n) = -8 - 1$
$3m = -9 \implies m = -3$.
Substituting $m = -3$ in $(i)$:
$-3 + 2n = 1 \implies 2n = 4 \implies n = 2$.
Thus,$(m, n) = (-3, 2)$.

(C) The vertices are $A(1, -1, 2)$,$B(2, 1, -1)$,and $C(3, -1, 2)$.
The position vectors are $\vec{a} = \hat{i} - \hat{j} + 2\hat{k}$,$\vec{b} = 2\hat{i} + \hat{j} - \hat{k}$,and $\vec{c} = 3\hat{i} - \hat{j} + 2\hat{k}$.
We find the vectors $\vec{AB}$ and $\vec{AC}$:
$\vec{AB} = \vec{b} - \vec{a} = (2-1)\hat{i} + (1 - (-1))\hat{j} + (-1-2)\hat{k} = \hat{i} + 2\hat{j} - 3\hat{k}$
$\vec{AC} = \vec{c} - \vec{a} = (3-1)\hat{i} + (-1 - (-1))\hat{j} + (2-2)\hat{k} = 2\hat{i} + 0\hat{j} + 0\hat{k}$
The area of the triangle is given by $\frac{1}{2} |\vec{AB} \times \vec{AC}|$.
$\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 2 & 0 & 0 \end{vmatrix} = \hat{i}(0 - 0) - \hat{j}(0 - (-6)) + \hat{k}(0 - 4) = -6\hat{j} - 4\hat{k}$.
The magnitude is $|\vec{AB} \times \vec{AC}| = \sqrt{(-6)^2 + (-4)^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13}$.
Therefore,the area of the triangle is $\frac{1}{2} \times 2\sqrt{13} = \sqrt{13}$ sq. units.

(C) Given $\overline{A} \times \overline{X} = \overline{B}$.
Taking the cross product with $\overline{A}$ on both sides:
$\overline{A} \times (\overline{A} \times \overline{X}) = \overline{A} \times \overline{B}$.
Using the vector triple product identity $\overline{a} \times (\overline{b} \times \overline{c}) = (\overline{a} \cdot \overline{c}) \overline{b} - (\overline{a} \cdot \overline{b}) \overline{c}$:
$(\overline{A} \cdot \overline{X}) \overline{A} - (\overline{A} \cdot \overline{A}) \overline{X} = \overline{A} \times \overline{B}$.
Substitute $\overline{A} \cdot \overline{X} = C$ and $\overline{A} \cdot \overline{A} = |\overline{A}|^2$:
$C \overline{A} - |\overline{A}|^2 \overline{X} = \overline{A} \times \overline{B}$.
Rearranging for $\overline{X}$:
$|\overline{A}|^2 \overline{X} = C \overline{A} - \overline{A} \times \overline{B}$.
$\overline{X} = \frac{C \overline{A} - \overline{A} \times \overline{B}}{|\overline{A}|^2}$.

(C) Given $\overline{R} \times \overline{B} = \overline{C} \times \overline{B}$,we can write $(\overline{R} - \overline{C}) \times \overline{B} = 0$.
This implies that $(\overline{R} - \overline{C})$ is parallel to $\overline{B}$,so $\overline{R} - \overline{C} = k\overline{B}$ for some scalar $k$.
Thus,$\overline{R} = \overline{C} + k\overline{B}$.
Given $\overline{R} \cdot \overline{A} = 0$,we substitute $\overline{R}$:
$(\overline{C} + k\overline{B}) \cdot \overline{A} = 0
\Rightarrow \overline{C} \cdot \overline{A} + k(\overline{B} \cdot \overline{A}) = 0$.
Calculate the dot products:
$\overline{A} \cdot \overline{C} = (2 \hat{i} + \hat{k}) \cdot (4 \hat{i} - 3 \hat{j} + 7 \hat{k}) = 2(4) + 0(-3) + 1(7) = 8 + 7 = 15$.
$\overline{A} \cdot \overline{B} = (2 \hat{i} + \hat{k}) \cdot (\hat{i} + \hat{j} + \hat{k}) = 2(1) + 0(1) + 1(1) = 2 + 1 = 3$.
Substituting these values:
$15 + k(3) = 0
\Rightarrow 3k = -15
\Rightarrow k = -5$.
Now,find $\overline{R}$:
$\overline{R} = \overline{C} - 5\overline{B} = (4 \hat{i} - 3 \hat{j} + 7 \hat{k}) - 5(\hat{i} + \hat{j} + \hat{k})
= (4-5)\hat{i} + (-3-5)\hat{j} + (7-5)\hat{k}
= -\hat{i} - 8\hat{j} + 2\hat{k}$.

(C) Given: $|\vec{a}| = \sqrt{27}$,$|\vec{b}| = 7$,and $|\vec{a} \times \vec{b}| = 35$.
We know that $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta$.
Therefore,$\sin \theta = \frac{|\vec{a} \times \vec{b}|}{|\vec{a}| |\vec{b}|} = \frac{35}{\sqrt{27} \times 7} = \frac{5}{\sqrt{27}}$.
Now,$\cos^2 \theta = 1 - \sin^2 \theta = 1 - \frac{25}{27} = \frac{2}{27}$.
Thus,$\cos \theta = \sqrt{\frac{2}{27}}$ (assuming $\theta$ is acute).
Finally,$\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta = \sqrt{27} \times 7 \times \sqrt{\frac{2}{27}} = 7 \sqrt{2}$.

(D) Given: $\overline{a}=2 \hat{i}+\hat{j}-2 \hat{k}$ and $\overline{b}=\hat{i}+\hat{j}$.
First,calculate the magnitude of $\overline{a}$:
$|\overline{a}|=\sqrt{2^2+1^2+(-2)^2}=\sqrt{4+1+4}=3$.
Next,calculate the cross product $\overline{a} \times \overline{b}$:
$\overline{a} \times \overline{b}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 1 & 1 & 0 \end{vmatrix} = \hat{i}(0 - (-2)) - \hat{j}(0 - (-2)) + \hat{k}(2 - 1) = 2 \hat{i}-2 \hat{j}+\hat{k}$.
The magnitude is $|\overline{a} \times \overline{b}|=\sqrt{2^2+(-2)^2+1^2}=\sqrt{4+4+1}=3$.
Given $|(\overline{a} \times \overline{b}) \times \overline{c}|=3$ and the angle $\theta$ between $\overline{c}$ and $\overline{a} \times \overline{b}$ is $30^{\circ}$.
Using the formula $|\overline{u} \times \overline{v}| = |\overline{u}||\overline{v}| \sin \theta$:
$|(\overline{a} \times \overline{b}) \times \overline{c}| = |\overline{a} \times \overline{b}| |\overline{c}| \sin 30^{\circ}$.
$3 = 3 \times |\overline{c}| \times \frac{1}{2} \Rightarrow |\overline{c}| = 2$.
Now,use the given condition $|\overline{c}-\overline{a}|=3$:
$|\overline{c}-\overline{a}|^2 = 3^2 = 9$.
$|\overline{c}|^2 + |\overline{a}|^2 - 2(\overline{a} \cdot \overline{c}) = 9$.
Substitute the known values $|\overline{c}|=2$ and $|\overline{a}|=3$:
$2^2 + 3^2 - 2(\overline{a} \cdot \overline{c}) = 9$.
$4 + 9 - 2(\overline{a} \cdot \overline{c}) = 9$.
$13 - 2(\overline{a} \cdot \overline{c}) = 9$.
$2(\overline{a} \cdot \overline{c}) = 4$.
$\overline{a} \cdot \overline{c} = 2$.

(C) Given that $\overline{a}, \overline{b}, \overline{c}$ are non-coplanar vectors,the scalar triple product $[\overline{a} \overline{b} \overline{c}] \neq 0$.
We know that $\overline{a} \cdot \overline{p} = \overline{a} \cdot \frac{\overline{b} \times \overline{c}}{[\overline{a} \overline{b} \overline{c}]} = \frac{[\overline{a} \overline{b} \overline{c}]}{[\overline{a} \overline{b} \overline{c}]} = 1$.
Similarly,$\overline{b} \cdot \overline{q} = \overline{b} \cdot \frac{\overline{c} \times \overline{a}}{[\overline{a} \overline{b} \overline{c}]} = \frac{[\overline{b} \overline{c} \overline{a}]}{[\overline{a} \overline{b} \overline{c}]} = 1$.
And $\overline{c} \cdot \overline{r} = \overline{c} \cdot \frac{\overline{a} \times \overline{b}}{[\overline{a} \overline{b} \overline{c}]} = \frac{[\overline{c} \overline{a} \overline{b}]}{[\overline{a} \overline{b} \overline{c}]} = 1$.
Substituting these values into the expression:
$2 \overline{a} \cdot \overline{p} + \overline{b} \cdot \overline{q} + \overline{c} \cdot \overline{r} = 2(1) + 1 + 1 = 4$.

(B) Given,$\bar{a}=\hat{i}-2 \hat{j}+3 \hat{k}$ and $\bar{b}=2 \hat{i}+3 \hat{j}-\hat{k}$.
First,calculate the vectors $(2 \bar{a}+\bar{b})$ and $(\bar{a}+2 \bar{b})$:
$2 \bar{a}+\bar{b} = 2(\hat{i}-2 \hat{j}+3 \hat{k}) + (2 \hat{i}+3 \hat{j}-\hat{k}) = (2+2)\hat{i} + (-4+3)\hat{j} + (6-1)\hat{k} = 4 \hat{i}-\hat{j}+5 \hat{k}$.
$\bar{a}+2 \bar{b} = (\hat{i}-2 \hat{j}+3 \hat{k}) + 2(2 \hat{i}+3 \hat{j}-\hat{k}) = (1+4)\hat{i} + (-2+6)\hat{j} + (3-2)\hat{k} = 5 \hat{i}+4 \hat{j}+\hat{k}$.
Now,calculate the magnitudes:
$|2 \bar{a}+\bar{b}| = \sqrt{4^2+(-1)^2+5^2} = \sqrt{16+1+25} = \sqrt{42}$.
$|\bar{a}+2 \bar{b}| = \sqrt{5^2+4^2+1^2} = \sqrt{25+16+1} = \sqrt{42}$.
Calculate the dot product:
$(2 \bar{a}+\bar{b}) \cdot (\bar{a}+2 \bar{b}) = (4)(5) + (-1)(4) + (5)(1) = 20 - 4 + 5 = 21$.
Using the formula $\cos \theta = \frac{(2 \bar{a}+\bar{b}) \cdot (\bar{a}+2 \bar{b})}{|2 \bar{a}+\bar{b}| |\bar{a}+2 \bar{b}|}$:
$\cos \theta = \frac{21}{\sqrt{42} \cdot \sqrt{42}} = \frac{21}{42} = \frac{1}{2}$.
Therefore,$\theta = \cos^{-1}(\frac{1}{2}) = \frac{\pi}{3}$.

(C) The area of a parallelogram with adjacent sides $\bar{a}$ and $\bar{b}$ is given by $|\bar{a} \times \bar{b}|$.
Given that $|\bar{a} \times \bar{b}| = 15$.
The area of the new parallelogram with adjacent sides $(3 \bar{a} + 2 \bar{b})$ and $(\bar{a} + 3 \bar{b})$ is given by the magnitude of their cross product:
$|(3 \bar{a} + 2 \bar{b}) \times (\bar{a} + 3 \bar{b})|$
$= |3(\bar{a} \times \bar{a}) + 9(\bar{a} \times \bar{b}) + 2(\bar{b} \times \bar{a}) + 6(\bar{b} \times \bar{b})|$
Since $\bar{a} \times \bar{a} = 0$ and $\bar{b} \times \bar{b} = 0$,and $\bar{b} \times \bar{a} = -(\bar{a} \times \bar{b})$,we have:
$= |0 + 9(\bar{a} \times \bar{b}) - 2(\bar{a} \times \bar{b}) + 0|$
$= |7(\bar{a} \times \bar{b})| = 7 |\bar{a} \times \bar{b}|$
$= 7 \times 15 = 105$ square units.

(C) Given,$\overline{a}=\hat{i}+\hat{j}+\hat{k}$ and $\overline{a} \times \overline{b}=\hat{j}-\hat{k}$.
Also,$\overline{a} \cdot \overline{b}=1$.
Let $\overline{b}=x \hat{i}+y \hat{j}+z \hat{k}$.
We know that $\overline{a} \times \overline{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ x & y & z \end{vmatrix} = (z-y)\hat{i} - (z-x)\hat{j} + (y-x)\hat{k}$.
Comparing this with $\hat{j}-\hat{k}$,we get:
$z-y=0 \implies z=y$ $(i)$
$-(z-x)=1 \implies x-z=1$ $(ii)$
$y-x=-1$ $(iii)$
From the dot product,$\overline{a} \cdot \overline{b} = x+y+z=1$ $(iv)$.
Substituting $z=y$ and $x=z+1$ into $(iv)$:
$(z+1) + z + z = 1 \implies 3z = 0 \implies z=0$.
Thus,$y=0$ and $x=0+1=1$.
Therefore,$\overline{b} = 1\hat{i} + 0\hat{j} + 0\hat{k} = \hat{i}$.

(B) First,calculate the cross product $\overline{a} \times \overline{b}$:
$\overline{a} \times \overline{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 1 & 1 & 0 \end{vmatrix} = \hat{i}(0 - (-2)) - \hat{j}(0 - (-2)) + \hat{k}(2 - 1) = 2 \hat{i} - 2 \hat{j} + \hat{k}$.
The magnitude is $|\overline{a} \times \overline{b}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4+4+1} = 3$ ...$(i)$.
Given $|\overline{c} - \overline{a}| = 2 \sqrt{2}$,squaring both sides gives $|\overline{c}|^2 + |\overline{a}|^2 - 2(\overline{c} \cdot \overline{a}) = 8$.
Since $|\overline{a}| = \sqrt{2^2 + 1^2 + (-2)^2} = 3$,we have $|\overline{a}|^2 = 9$.
Given $\overline{a} \cdot \overline{c} = |\overline{c}|$,the equation becomes $|\overline{c}|^2 + 9 - 2|\overline{c}| = 8$.
This simplifies to $|\overline{c}|^2 - 2|\overline{c}| + 1 = 0$,which is $(|\overline{c}| - 1)^2 = 0$,so $|\overline{c}| = 1$ ...$(ii)$.
The magnitude of the cross product is $|(\overline{a} \times \overline{b}) \times \overline{c}| = |\overline{a} \times \overline{b}| |\overline{c}| \sin(60^{\circ})$.
Substituting the values: $(3)(1)(\frac{\sqrt{3}}{2}) = \frac{3 \sqrt{3}}{2}$.

(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) Let $\theta$ be the angle between $\hat{a}$ and $\hat{b}$.
Since $\bar{c}=\hat{a}+2 \hat{b}$ and $\bar{d}=5 \hat{a}-4 \hat{b}$ are perpendicular,their dot product is zero.
$\bar{c} \cdot \bar{d} = 0$
$(\hat{a}+2 \hat{b}) \cdot (5 \hat{a}-4 \hat{b}) = 0$
$5(\hat{a} \cdot \hat{a}) - 4(\hat{a} \cdot \hat{b}) + 10(\hat{b} \cdot \hat{a}) - 8(\hat{b} \cdot \hat{b}) = 0$
Since $\hat{a}$ and $\hat{b}$ are unit vectors,$|\hat{a}|=1$ and $|\hat{b}|=1$,and $\hat{a} \cdot \hat{a} = 1$,$\hat{b} \cdot \hat{b} = 1$.
$5(1) + 6(\hat{a} \cdot \hat{b}) - 8(1) = 0$
$6(\hat{a} \cdot \hat{b}) - 3 = 0$
$6 \cos \theta = 3$
$\cos \theta = \frac{3}{6} = \frac{1}{2}$
$\theta = \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3}$

(B) The area of a parallelogram with adjacent sides $\overline{a}$ and $\overline{b}$ is given by $|\overline{a} \times \overline{b}|$.
First,we calculate the cross product $\overline{a} \times \overline{b}$:
$\overline{a} \times \overline{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -\alpha & 1 \\ 1 & \alpha & 3 \end{vmatrix} = \hat{i}(-3\alpha - \alpha) - \hat{j}(9 - 1) + \hat{k}(3\alpha + \alpha) = -4\alpha \hat{i} - 8 \hat{j} + 4\alpha \hat{k}$.
The magnitude is $|\overline{a} \times \overline{b}| = \sqrt{(-4\alpha)^2 + (-8)^2 + (4\alpha)^2} = \sqrt{16\alpha^2 + 64 + 16\alpha^2} = \sqrt{32\alpha^2 + 64}$.
Given the area is $8\sqrt{3}$,we have $\sqrt{32\alpha^2 + 64} = 8\sqrt{3}$.
Squaring both sides: $32\alpha^2 + 64 = 64 \times 3 = 192$.
$32\alpha^2 = 128 \Rightarrow \alpha^2 = 4$.
Now,calculate the dot product $\overline{a} \cdot \overline{b} = (3\hat{i} - \alpha\hat{j} + \hat{k}) \cdot (\hat{i} + \alpha\hat{j} + 3\hat{k}) = 3(1) - \alpha(\alpha) + 1(3) = 3 - \alpha^2 + 3 = 6 - \alpha^2$.
Substituting $\alpha^2 = 4$,we get $\overline{a} \cdot \overline{b} = 6 - 4 = 2$.

(B) Given that $|\overline{a}| = 2, |\overline{b}| = 3, |\overline{c}| = 4$.
Since $\overline{a} \perp (\overline{b}+\overline{c})$,we have $\overline{a} \cdot (\overline{b}+\overline{c}) = 0 \Rightarrow \overline{a} \cdot \overline{b} + \overline{a} \cdot \overline{c} = 0$.
Since $\overline{b} \perp (\overline{c}+\overline{a})$,we have $\overline{b} \cdot (\overline{c}+\overline{a}) = 0 \Rightarrow \overline{b} \cdot \overline{c} + \overline{b} \cdot \overline{a} = 0$.
Since $\overline{c} \perp (\overline{a}+\overline{b})$,we have $\overline{c} \cdot (\overline{a}+\overline{b}) = 0 \Rightarrow \overline{c} \cdot \overline{a} + \overline{c} \cdot \overline{b} = 0$.
Adding these three equations:
$2(\overline{a} \cdot \overline{b} + \overline{b} \cdot \overline{c} + \overline{c} \cdot \overline{a}) = 0 \Rightarrow \overline{a} \cdot \overline{b} + \overline{b} \cdot \overline{c} + \overline{c} \cdot \overline{a} = 0$.
Now,consider the magnitude squared of the sum:
$|\overline{a}+\overline{b}+\overline{c}|^2 = |\overline{a}|^2 + |\overline{b}|^2 + |\overline{c}|^2 + 2(\overline{a} \cdot \overline{b} + \overline{b} \cdot \overline{c} + \overline{c} \cdot \overline{a})$.
Substituting the known values:
$|\overline{a}+\overline{b}+\overline{c}|^2 = 2^2 + 3^2 + 4^2 + 2(0) = 4 + 9 + 16 = 29$.
Therefore,$|\overline{a}+\overline{b}+\overline{c}| = \sqrt{29}$.

(D) Given that $|\overline{A}|=3, |\overline{B}|=4, |\overline{C}|=5$.
Since $\overline{A} \perp (\overline{B}+\overline{C})$,we have $\overline{A} \cdot (\overline{B}+\overline{C}) = 0 \Rightarrow \overline{A} \cdot \overline{B} + \overline{A} \cdot \overline{C} = 0$.
Since $\overline{B} \perp (\overline{C}+\overline{A})$,we have $\overline{B} \cdot (\overline{C}+\overline{A}) = 0 \Rightarrow \overline{B} \cdot \overline{C} + \overline{B} \cdot \overline{A} = 0$.
Since $\overline{C} \perp (\overline{A}+\overline{B})$,we have $\overline{C} \cdot (\overline{A}+\overline{B}) = 0 \Rightarrow \overline{C} \cdot \overline{A} + \overline{C} \cdot \overline{B} = 0$.
Adding these three equations,we get $2(\overline{A} \cdot \overline{B} + \overline{B} \cdot \overline{C} + \overline{C} \cdot \overline{A}) = 0$.
Now,consider $|\overline{A}+\overline{B}+\overline{C}|^2 = |\overline{A}|^2 + |\overline{B}|^2 + |\overline{C}|^2 + 2(\overline{A} \cdot \overline{B} + \overline{B} \cdot \overline{C} + \overline{C} \cdot \overline{A})$.
Substituting the values,$|\overline{A}+\overline{B}+\overline{C}|^2 = 3^2 + 4^2 + 5^2 + 0 = 9 + 16 + 25 = 50$.
Therefore,$|\overline{A}+\overline{B}+\overline{C}| = \sqrt{50} = 5 \sqrt{2}$.

(B) Let the vertices of the triangle be $O(0), A(\bar{a}+2\bar{b}), B(\bar{a}-2\bar{b})$.
The area of the triangle is given by $\frac{1}{2} |\vec{OA} \times \vec{OB}|$.
$\vec{OA} = \bar{a}+2\bar{b}$ and $\vec{OB} = \bar{a}-2\bar{b}$.
$\vec{OA} \times \vec{OB} = (\bar{a}+2\bar{b}) \times (\bar{a}-2\bar{b})$
$= \bar{a} \times \bar{a} - 2(\bar{a} \times \bar{b}) + 2(\bar{b} \times \bar{a}) - 4(\bar{b} \times \bar{b})$
Since $\bar{a} \times \bar{a} = 0$ and $\bar{b} \times \bar{b} = 0$,and $\bar{b} \times \bar{a} = -(\bar{a} \times \bar{b})$:
$= 0 - 2(\bar{a} \times \bar{b}) - 2(\bar{a} \times \bar{b}) - 0 = -4(\bar{a} \times \bar{b})$.
Area $= \frac{1}{2} |-4(\bar{a} \times \bar{b})| = 2 |\bar{a} \times \bar{b}|$.
Since $\bar{a} \perp \bar{b}$,$|\bar{a} \times \bar{b}| = |\bar{a}| |\bar{b}| \sin(90^{\circ}) = 2 \times 3 \times 1 = 6$.
Area $= 2 \times 6 = 12 \text{ sq. units}$.

(B) Given,$|\bar{a}|=1, |\bar{b}|=1$ and $|\bar{c}|=2$.
Using the vector triple product formula $\bar{a} \times (\bar{a} \times \bar{c}) = (\bar{a} \cdot \bar{c})\bar{a} - (\bar{a} \cdot \bar{a})\bar{c}$.
Given equation: $(\bar{a} \cdot \bar{c})\bar{a} - |\bar{a}|^2\bar{c} + \bar{b} = \bar{0}$.
Since $|\bar{a}|=1$,we have $(\bar{a} \cdot \bar{c})\bar{a} - \bar{c} = -\bar{b}$.
Taking the magnitude squared on both sides: $|(\bar{a} \cdot \bar{c})\bar{a} - \bar{c}|^2 = |-\bar{b}|^2$.
$(\bar{a} \cdot \bar{c})^2 |\bar{a}|^2 + |\bar{c}|^2 - 2(\bar{a} \cdot \bar{c})(\bar{a} \cdot \bar{c}) = |\bar{b}|^2$.
$(\bar{a} \cdot \bar{c})^2(1) + 4 - 2(\bar{a} \cdot \bar{c})^2 = 1$.
$-(\bar{a} \cdot \bar{c})^2 = -3 \Rightarrow (\bar{a} \cdot \bar{c})^2 = 3$.
Thus,$\bar{a} \cdot \bar{c} = \sqrt{3}$ (for acute angle).
$|\bar{a}||\bar{c}| \cos \theta = \sqrt{3} \Rightarrow (1)(2) \cos \theta = \sqrt{3}$.
$\cos \theta = \frac{\sqrt{3}}{2} \Rightarrow \theta = \frac{\pi}{6}$.

(D) Given that $\overline{b}+\lambda \overline{a}$ is perpendicular to $\overline{c}$,their dot product must be zero: $(\overline{b}+\lambda \overline{a}) \cdot \overline{c} = 0$.
First,calculate $\overline{b}+\lambda \overline{a}$:
$\overline{b}+\lambda \overline{a} = (-\hat{i}+2 \hat{j}+\hat{k}) + \lambda(2 \hat{i}+2 \hat{j}+3 \hat{k}) = (-1+2\lambda)\hat{i} + (2+2\lambda)\hat{j} + (1+3\lambda)\hat{k}$.
Now,take the dot product with $\overline{c} = 3 \hat{i}+\hat{j}$:
$((-1+2\lambda)\hat{i} + (2+2\lambda)\hat{j} + (1+3\lambda)\hat{k}) \cdot (3 \hat{i}+\hat{j}) = 0$.
$3(-1+2\lambda) + 1(2+2\lambda) + 0(1+3\lambda) = 0$.
$-3 + 6\lambda + 2 + 2\lambda = 0$.
$8\lambda - 1 = 0$.
$8\lambda = 1$.
$\lambda = \frac{1}{8}$.

(B) Given: $\bar{a} = \hat{j} - \hat{k}$ and $\bar{c} = \hat{i} - \hat{j} - \hat{k}$.
Let $\bar{b} = x\hat{i} + y\hat{j} + z\hat{k}$.
From $\bar{a} \times \bar{b} + \bar{c} = \vec{0}$,we have $\bar{a} \times \bar{b} = -\bar{c} = -\hat{i} + \hat{j} + \hat{k}$.
Calculating $\bar{a} \times \bar{b}$:
$\bar{a} \times \bar{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & -1 \\ x & y & z \end{vmatrix} = \hat{i}(z + y) - \hat{j}(z) + \hat{k}(-x) = (y + z)\hat{i} - z\hat{j} - x\hat{k}$.
Comparing with $-\hat{i} + \hat{j} + \hat{k}$:
$y + z = -1$,$-z = 1 \Rightarrow z = -1$,$-x = 1 \Rightarrow x = -1$.
Substituting $z = -1$ into $y + z = -1$,we get $y - 1 = -1 \Rightarrow y = 0$.
Now,check $\bar{a} \cdot \bar{b} = 3$:
$\bar{a} \cdot \bar{b} = (0\hat{i} + 1\hat{j} - 1\hat{k}) \cdot (x\hat{i} + y\hat{j} + z\hat{k}) = y - z = 0 - (-1) = 1$.
Since $1 \neq 3$,there is a contradiction in the problem statement's constraints. However,solving for $\bar{b}$ using the cross product condition gives $\bar{b} = -\hat{i} + 0\hat{j} - \hat{k}$. Re-evaluating the system: if $\bar{a} \cdot \bar{b} = 3$ must hold,the vector $\bar{b}$ is not uniquely determined by these equations. Given the options,option $B$ is $-\hat{i} + \hat{j} - 2\hat{k}$. Let's check: $\bar{a} \cdot \bar{b} = 1 - (-2) = 3$ and $\bar{a} \times \bar{b} = (1 - 2)\hat{i} - (-2)\hat{j} - (-1)\hat{k} = -\hat{i} + 2\hat{j} + \hat{k} \neq -\bar{c}$. The provided options do not satisfy both conditions simultaneously.

(D) The given lines are $\bar{r} = \bar{a}_1 + \lambda \bar{b}_1$ and $\bar{r} = \bar{a}_2 + \mu \bar{b}_2$.
Here,$\bar{a}_1 = \hat{i} + 2\hat{j} - \hat{k}$,$\bar{b}_1 = 2\hat{i} + \hat{j} - 3\hat{k}$,$\bar{a}_2 = 2\hat{i} - \hat{j} + 2\hat{k}$,and $\bar{b}_2 = \hat{i} - \hat{j} + \hat{k}$.
First,calculate $\bar{a}_2 - \bar{a}_1 = (2-1)\hat{i} + (-1-2)\hat{j} + (2-(-1))\hat{k} = \hat{i} - 3\hat{j} + 3\hat{k}$.
Next,calculate the cross product $\bar{b}_1 \times \bar{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -3 \\ 1 & -1 & 1 \end{vmatrix} = \hat{i}(1-3) - \hat{j}(2+3) + \hat{k}(-2-1) = -2\hat{i} - 5\hat{j} - 3\hat{k}$.
The magnitude is $|\bar{b}_1 \times \bar{b}_2| = \sqrt{(-2)^2 + (-5)^2 + (-3)^2} = \sqrt{4 + 25 + 9} = \sqrt{38}$.
The shortest distance is $d = \left| \frac{(\bar{b}_1 \times \bar{b}_2) \cdot (\bar{a}_2 - \bar{a}_1)}{|\bar{b}_1 \times \bar{b}_2|} \right| = \left| \frac{(-2\hat{i} - 5\hat{j} - 3\hat{k}) \cdot (\hat{i} - 3\hat{j} + 3\hat{k})}{\sqrt{38}} \right| = \left| \frac{-2 + 15 - 9}{\sqrt{38}} \right| = \frac{4}{\sqrt{38}} = \frac{4}{\sqrt{2} \times \sqrt{19}} = \frac{2 \times 2}{\sqrt{2} \times \sqrt{19}} = \frac{2 \sqrt{2}}{\sqrt{19}}$ units.

(C) Let the required vector be $\vec{r}$. Since $\vec{r}$ is perpendicular to $\vec{a} = 2\hat{i} + \hat{j} - 3\hat{k}$ and $\vec{b} = \hat{i} - 2\hat{j} + \hat{k}$,it must be parallel to $\vec{a} \times \vec{b}$.
First,calculate the cross product $\vec{a} \times \vec{b}$:
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -3 \\ 1 & -2 & 1 \end{vmatrix} = \hat{i}(1 - 6) - \hat{j}(2 + 3) + \hat{k}(-4 - 1) = -5\hat{i} - 5\hat{j} - 5\hat{k} = -5(\hat{i} + \hat{j} + \hat{k})$.
The unit vector perpendicular to both is $\hat{n} = \pm \frac{-5(\hat{i} + \hat{j} + \hat{k})}{|-5(\hat{i} + \hat{j} + \hat{k})|} = \pm \frac{-5(\hat{i} + \hat{j} + \hat{k})}{5\sqrt{3}} = \mp \frac{1}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k})$.
The required vector of magnitude $6$ is $\vec{r} = 6 \times \hat{n} = \pm \frac{6}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k}) = \pm 2\sqrt{3}(\hat{i} + \hat{j} + \hat{k})$.
Comparing with the given options,the correct vector is $2\sqrt{3}(\hat{i} + \hat{j} + \hat{k})$.

(A) Since the vectors $\overline{a}$,$\overline{b}$,and $\overline{c}$ are mutually orthogonal,their dot products must be zero: $\overline{a} \cdot \overline{c} = 0$ and $\overline{b} \cdot \overline{c} = 0$.
For $\overline{a} \cdot \overline{c} = 0$:
$(\hat{i}-\hat{j}+2 \hat{k}) \cdot (\lambda \hat{i}+\hat{j}+\mu \hat{k}) = 0$
$\lambda - 1 + 2\mu = 0 \implies \lambda + 2\mu = 1 \quad ...(i)$
For $\overline{b} \cdot \overline{c} = 0$:
$(2 \hat{i}+4 \hat{j}+\hat{k}) \cdot (\lambda \hat{i}+\hat{j}+\mu \hat{k}) = 0$
$2\lambda + 4 + \mu = 0 \implies 2\lambda + \mu = -4 \quad ...(ii)$
Multiplying equation $(ii)$ by $2$,we get $4\lambda + 2\mu = -8 \quad ...(iii)$.
Subtracting $(i)$ from $(iii)$:
$(4\lambda + 2\mu) - (\lambda + 2\mu) = -8 - 1$
$3\lambda = -9 \implies \lambda = -3$.
Substituting $\lambda = -3$ into $(i)$:
$-3 + 2\mu = 1
2\mu = 4
\mu = 2$.
Thus,$(\lambda, \mu) = (-3, 2)$.

(B) The vector perpendicular to $\overline{a}$ and $\overline{b}$ is given by the cross product $\overline{a} \times \overline{b}$.
$\overline{a} \times \overline{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{vmatrix} = \hat{i}(1-0) - \hat{j}(1-0) + \hat{k}(1-0) = \hat{i} - \hat{j} + \hat{k}$.
The magnitude of this vector is $|\overline{a} \times \overline{b}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}$.
The unit vectors perpendicular to both $\overline{a}$ and $\overline{b}$ are given by $\pm \frac{\overline{a} \times \overline{b}}{|\overline{a} \times \overline{b}|} = \pm \frac{1}{\sqrt{3}}(\hat{i} - \hat{j} + \hat{k})$.
Thus,there are exactly two such unit vectors.

(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) Given $\bar{u} \cdot \hat{n}=0$ and $\bar{v} \cdot \hat{n}=0$.
This implies that the unit vector $\hat{n}$ is perpendicular to both $\bar{u}$ and $\bar{v}$.
Therefore,$\hat{n}$ is parallel to the cross product $\bar{u} \times \bar{v}$.
Thus,$\hat{n} = \pm \frac{\bar{u} \times \bar{v}}{|\bar{u} \times \bar{v}|}$.
First,calculate the cross product:
$\bar{u} \times \bar{v} = (\hat{i}+\hat{j}) \times (\hat{i}-\hat{j}) = \hat{i} \times \hat{i} - \hat{i} \times \hat{j} + \hat{j} \times \hat{i} - \hat{j} \times \hat{j} = 0 - \hat{k} - \hat{k} - 0 = -2\hat{k}$.
The magnitude is $|\bar{u} \times \bar{v}| = |-2\hat{k}| = 2$.
So,$\hat{n} = \pm \frac{-2\hat{k}}{2} = \pm \hat{k}$.
Now,calculate $|\bar{w} \cdot \hat{n}|$:
$|\bar{w} \cdot \hat{n}| = |(\hat{i}+2\hat{j}+3\hat{k}) \cdot (\pm \hat{k})| = |\pm 3| = 3$.

(C) Since the vectors $\bar{a}$,$\bar{b}$,and $\bar{c}$ are mutually orthogonal,their dot products must be zero.
First,check $\bar{a} \cdot \bar{b} = (1)(2) + (-1)(4) + (2)(1) = 2 - 4 + 2 = 0$.
Next,use $\bar{a} \cdot \bar{c} = 0$:
$(1)(p) + (-1)(1) + (2)(q) = p - 1 + 2q = 0 \implies p + 2q = 1$ ... $(i)$
Then,use $\bar{b} \cdot \bar{c} = 0$:
$(2)(p) + (4)(1) + (1)(q) = 2p + 4 + q = 0 \implies 2p + q = -4$ ... $(ii)$
From $(ii)$,$q = -4 - 2p$. Substituting this into $(i)$:
$p + 2(-4 - 2p) = 1$
$p - 8 - 4p = 1$
$-3p = 9 \implies p = -3$.
Substituting $p = -3$ into $q = -4 - 2p$:
$q = -4 - 2(-3) = -4 + 6 = 2$.
Thus,$(p, q) = (-3, 2)$.

(B) Let $\theta$ be the angle between $\overline{a}$ and $\overline{b}$.
Since $\overline{a}$ and $\overline{b}$ are unit vectors,$|\overline{a}| = 1$ and $|\overline{b}| = 1$.
Given that $(5 \overline{a} + 4 \overline{b})$ and $(\overline{a} - 2 \overline{b})$ are perpendicular,their dot product is zero.
$(5 \overline{a} + 4 \overline{b}) \cdot (\overline{a} - 2 \overline{b}) = 0$
$5(\overline{a} \cdot \overline{a}) - 10(\overline{a} \cdot \overline{b}) + 4(\overline{b} \cdot \overline{a}) - 8(\overline{b} \cdot \overline{b}) = 0$
$5|\overline{a}|^2 - 6(\overline{a} \cdot \overline{b}) - 8|\overline{b}|^2 = 0$
Since $|\overline{a}| = 1$ and $|\overline{b}| = 1$,we have $5(1)^2 - 6(1)(1)\cos \theta - 8(1)^2 = 0$.
$5 - 6 \cos \theta - 8 = 0$
$-6 \cos \theta = 3$
$\cos \theta = -\frac{3}{6} = -\frac{1}{2}$
Since $\cos \theta = -\frac{1}{2}$,the angle $\theta = \frac{2\pi}{3}$.

(B) Let $\theta$ be the angle between $\overline{a}$ and $\overline{b}$.
Since $\overline{a}$ and $\overline{b}$ are unit vectors,$|\overline{a}| = 1$ and $|\overline{b}| = 1$.
Given that $\overline{c} = \overline{a} + 2\overline{b}$ and $\overline{d} = 5\overline{a} - 4\overline{b}$ are perpendicular,their dot product is zero:
$\overline{c} \cdot \overline{d} = 0$
$(\overline{a} + 2\overline{b}) \cdot (5\overline{a} - 4\overline{b}) = 0$
$5(\overline{a} \cdot \overline{a}) - 4(\overline{a} \cdot \overline{b}) + 10(\overline{b} \cdot \overline{a}) - 8(\overline{b} \cdot \overline{b}) = 0$
$5|\overline{a}|^2 + 6(\overline{a} \cdot \overline{b}) - 8|\overline{b}|^2 = 0$
Since $|\overline{a}| = 1$ and $|\overline{b}| = 1$,and $\overline{a} \cdot \overline{b} = |\overline{a}||\overline{b}| \cos \theta = \cos \theta$:
$5(1) + 6 \cos \theta - 8(1) = 0$
$6 \cos \theta - 3 = 0$
$6 \cos \theta = 3$
$\cos \theta = \frac{3}{6} = \frac{1}{2}$
$\theta = \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3}$.

(C) Let $\overline{u} = 3 \overline{a} + 5 \overline{b}$ and $\overline{v} = 5 \overline{a} + 3 \overline{b}$.
$\overline{u} = 3(\hat{i}-2 \hat{j}+3 \hat{k}) + 5(2 \hat{i}+3 \hat{j}-\hat{k}) = (3+10)\hat{i} + (-6+15)\hat{j} + (9-5)\hat{k} = 13 \hat{i} + 9 \hat{j} + 4 \hat{k}$.
$\overline{v} = 5(\hat{i}-2 \hat{j}+3 \hat{k}) + 3(2 \hat{i}+3 \hat{j}-\hat{k}) = (5+6)\hat{i} + (-10+9)\hat{j} + (15-3)\hat{k} = 11 \hat{i} - \hat{j} + 12 \hat{k}$.
The angle $\theta$ between $\overline{u}$ and $\overline{v}$ is given by $\cos \theta = \frac{\overline{u} \cdot \overline{v}}{|\overline{u}| |\overline{v}|}$.
$\overline{u} \cdot \overline{v} = (13)(11) + (9)(-1) + (4)(12) = 143 - 9 + 48 = 182$.
$|\overline{u}| = \sqrt{13^2 + 9^2 + 4^2} = \sqrt{169 + 81 + 16} = \sqrt{266}$.
$|\overline{v}| = \sqrt{11^2 + (-1)^2 + 12^2} = \sqrt{121 + 1 + 144} = \sqrt{266}$.
$\cos \theta = \frac{182}{\sqrt{266} \cdot \sqrt{266}} = \frac{182}{266} = \frac{13}{19}$.
Therefore,$\theta = \cos^{-1}\left(\frac{13}{19}\right)$.

(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:
$= 0 + 0 = 0$

(B) Given $|\overline{c}-\overline{a}|=2 \sqrt{2}$. Squaring both sides,we get $|\overline{c}|^2+|\overline{a}|^2-2(\overline{a} \cdot \overline{c})=8$.
Since $|\overline{a}|^2 = 2^2+1^2+(-2)^2 = 9$ and $\overline{a} \cdot \overline{c}=|\overline{c}|$,the equation becomes $|\overline{c}|^2+9-2|\overline{c}|=8$.
This simplifies to $(|\overline{c}|-1)^2=0$,so $|\overline{c}|=1$.
Next,calculate $\overline{a} \times \overline{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 1 & 1 & 0 \end{vmatrix} = \hat{i}(0 - (-2)) - \hat{j}(0 - (-2)) + \hat{k}(2 - 1) = 2\hat{i}-2\hat{j}+\hat{k}$.
The magnitude is $|\overline{a} \times \overline{b}| = \sqrt{2^2+(-2)^2+1^2} = \sqrt{4+4+1} = 3$.
Finally,$|(\overline{a} \times \overline{b}) \times \overline{c}| = |\overline{a} \times \overline{b}| |\overline{c}| \sin(30^{\circ}) = 3 \times 1 \times \frac{1}{2} = \frac{3}{2}$.

(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.