MHT CET 2023 Mathematics Question Paper with Answer and Solution

(C) Given $|\overline{a}| = \sqrt{3}, |\overline{b}| = 1, |\overline{c}| = 2$.
Using the vector triple product formula $\overline{a} \times (\overline{a} \times \overline{c}) = (\overline{a} \cdot \overline{c}) \overline{a} - (\overline{a} \cdot \overline{a}) \overline{c}$.
Substituting this into the given equation: $(\overline{a} \cdot \overline{c}) \overline{a} - |\overline{a}|^2 \overline{c} + 3 \overline{b} = \overline{0}$.
Since $|\overline{a}|^2 = (\sqrt{3})^2 = 3$,we have $(\overline{a} \cdot \overline{c}) \overline{a} - 3 \overline{c} = -3 \overline{b}$.
Taking the squared magnitude on both sides: $|(\overline{a} \cdot \overline{c}) \overline{a} - 3 \overline{c}|^2 = |-3 \overline{b}|^2$.
Expanding the left side: $(\overline{a} \cdot \overline{c})^2 |\overline{a}|^2 + 9 |\overline{c}|^2 - 6 (\overline{a} \cdot \overline{c})(\overline{a} \cdot \overline{c}) = 9 |\overline{b}|^2$.
$(\overline{a} \cdot \overline{c})^2 (3) + 9 (2)^2 - 6 (\overline{a} \cdot \overline{c})^2 = 9 (1)^2$.
$-3 (\overline{a} \cdot \overline{c})^2 + 36 = 9$.
$-3 (\overline{a} \cdot \overline{c})^2 = -27 \implies (\overline{a} \cdot \overline{c})^2 = 9$.
Thus,$\overline{a} \cdot \overline{c} = \pm 3$.
Since $\overline{a} \cdot \overline{c} = |\overline{a}| |\overline{c}| \cos \theta$,we have $(\sqrt{3})(2) \cos \theta = \pm 3$.
$\cos \theta = \pm \frac{3}{2\sqrt{3}} = \pm \frac{\sqrt{3}}{2}$.
Therefore,$\sec^2 \theta = \frac{1}{\cos^2 \theta} = \frac{1}{3/4} = \frac{4}{3}$.

(B) Given that $\overline{a}, \overline{b}, \overline{c}$ are unit vectors,so $|\overline{a}| = |\overline{b}| = |\overline{c}| = 1$.
Given equation: $\overline{a} + 2\overline{c} = -2\overline{b}$.
Squaring both sides:
$|\overline{a} + 2\overline{c}|^2 = |-2\overline{b}|^2$
$|\overline{a}|^2 + 4|\overline{c}|^2 + 4(\overline{a} \cdot \overline{c}) = 4|\overline{b}|^2$
Since $|\overline{a}| = |\overline{b}| = |\overline{c}| = 1$,we have:
$1 + 4(1) + 4(\overline{a} \cdot \overline{c}) = 4(1)$
$5 + 4(\overline{a} \cdot \overline{c}) = 4$
$4(\overline{a} \cdot \overline{c}) = -1$
$\overline{a} \cdot \overline{c} = -\frac{1}{4}$
Since $\overline{a} \cdot \overline{c} = |\overline{a}||\overline{c}| \cos \theta = \cos \theta$,we have $\cos \theta = -\frac{1}{4}$.
Then $\sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - (-\frac{1}{4})^2} = \sqrt{1 - \frac{1}{16}} = \sqrt{\frac{15}{16}} = \frac{\sqrt{15}}{4}$.
Finally,$|\overline{a} \times \overline{c}| = |\overline{a}||\overline{c}| \sin \theta = (1)(1) \frac{\sqrt{15}}{4} = \frac{\sqrt{15}}{4}$.

(D) Given $\overline{b} \cdot \overline{c} = 10$.
Since $\overline{b} \cdot \overline{c} = |\overline{b}| |\overline{c}| \cos(\frac{\pi}{3}) = 10$,we have $(5) |\overline{c}| (\frac{1}{2}) = 10$,which implies $|\overline{c}| = 4$.
We are given that $\overline{a}$ is perpendicular to $\overline{b} \times \overline{c}$,so the angle between $\overline{a}$ and $\overline{b} \times \overline{c}$ is $\frac{\pi}{2}$.
Therefore,$|\overline{a} \times (\overline{b} \times \overline{c})| = |\overline{a}| |\overline{b} \times \overline{c}| \sin(\frac{\pi}{2}) = |\overline{a}| |\overline{b} \times \overline{c}|$.
Now,$|\overline{b} \times \overline{c}| = |\overline{b}| |\overline{c}| \sin(\frac{\pi}{3}) = (5)(4)(\frac{\sqrt{3}}{2}) = 10\sqrt{3}$.
Thus,$|\overline{a} \times (\overline{b} \times \overline{c})| = (\sqrt{3})(10\sqrt{3}) = 10 \times 3 = 30$.

(B) Given: $|\overline{a}|=3$,$|\overline{b}|=5$,$\overline{b} \cdot \overline{c}=10$,and the angle $\theta$ between $\overline{b}$ and $\overline{c}$ is $\frac{\pi}{3}$.
First,find $|\overline{c}|$: $\overline{b} \cdot \overline{c} = |\overline{b}| |\overline{c}| \cos(\frac{\pi}{3}) = 10$.
$5 \times |\overline{c}| \times \frac{1}{2} = 10 \implies |\overline{c}| = 4$.
Next,find $|\overline{b} \times \overline{c}|$: $|\overline{b} \times \overline{c}| = |\overline{b}| |\overline{c}| \sin(\frac{\pi}{3}) = 5 \times 4 \times \frac{\sqrt{3}}{2} = 10\sqrt{3}$.
Since $\overline{a}$ is perpendicular to $\overline{b} \times \overline{c}$,the angle between $\overline{a}$ and $(\overline{b} \times \overline{c})$ is $\frac{\pi}{2}$.
Therefore,$|\overline{a} \times (\overline{b} \times \overline{c})| = |\overline{a}| |\overline{b} \times \overline{c}| \sin(\frac{\pi}{2}) = 3 \times 10\sqrt{3} \times 1 = 30\sqrt{3}$.
Wait,re-evaluating the options provided,there might be a typo in the question or options. Given the standard interpretation,the result is $30\sqrt{3}$. If we assume the question implies $\overline{a}$ is in the plane of $\overline{b}$ and $\overline{c}$ or similar,the result changes. However,based on the provided perpendicular condition,the calculation holds.

(B) The function $f(x)$ is defined as the scalar triple product of vectors $\overline{a}, \overline{b}, \text{ and } \overline{c}$:
$f(x) = \begin{vmatrix} x & -2 & 3 \\ -2 & x & -1 \\ 7 & -2 & x \end{vmatrix}$
Expanding the determinant along the first row:
$f(x) = x(x^2 - 2) - (-2)(-2x + 7) + 3(4 - 7x)$
$f(x) = x^3 - 2x + 4x - 14 + 12 - 21x$
$f(x) = x^3 - 19x - 2$ (Note: Correcting the expansion: $x(x^2-2) + 2(-2x+7) + 3(4-7x) = x^3 - 2x - 4x + 14 + 12 - 21x = x^3 - 27x + 26$)
To find the local minima,we find the critical points by setting $f'(x) = 0$:
$f'(x) = 3x^2 - 27 = 0$
$3(x^2 - 9) = 0 \Rightarrow x = \pm 3$
Using the second derivative test,$f''(x) = 6x$:
At $x = 3$,$f''(3) = 18 > 0$,so $x_0 = 3$ is the point of local minima.
At $x = 3$,$\overline{a} = 3 \hat{i} - 2 \hat{j} + 3 \hat{k}$ and $\overline{b} = -2 \hat{i} + 3 \hat{j} - \hat{k}$.
Calculating the dot product $\overline{a} \cdot \overline{b}$:
$\overline{a} \cdot \overline{b} = (3)(-2) + (-2)(3) + (3)(-1) = -6 - 6 - 3 = -15$.

(D) Given,$\overline{b} \times \overline{c} = \overline{b} \times \overline{a}$.
This implies $\overline{b} \times (\overline{c} - \overline{a}) = \overline{0}$.
Thus,$\overline{c} - \overline{a} = \lambda \overline{b}$ for some scalar $\lambda$,so $\overline{c} = \overline{a} + \lambda \overline{b}$.
Given $\overline{c} \cdot \overline{a} = 0$,we have $(\overline{a} + \lambda \overline{b}) \cdot \overline{a} = 0$.
This gives $|\overline{a}|^2 + \lambda (\overline{b} \cdot \overline{a}) = 0$.
Calculating the values: $|\overline{a}|^2 = 1^2 + 2^2 + (-1)^2 = 6$ and $\overline{b} \cdot \overline{a} = (1)(1) + (1)(2) + (-1)(-1) = 1 + 2 + 1 = 4$.
Substituting these: $6 + 4\lambda = 0 \Rightarrow \lambda = -\frac{6}{4} = -\frac{3}{2}$.
Now,$\overline{c} = \overline{a} - \frac{3}{2} \overline{b}$.
Then $\overline{c} \cdot \overline{b} = (\overline{a} - \frac{3}{2} \overline{b}) \cdot \overline{b} = \overline{a} \cdot \overline{b} - \frac{3}{2} |\overline{b}|^2$.
We have $\overline{a} \cdot \overline{b} = 4$ and $|\overline{b}|^2 = 1^2 + 1^2 + (-1)^2 = 3$.
Therefore,$\overline{c} \cdot \overline{b} = 4 - \frac{3}{2}(3) = 4 - \frac{9}{2} = -\frac{1}{2}$.

(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 $|\overline{a}| = \sqrt{2^2+1^2+(-2)^2} = \sqrt{9} = 3$ and $\overline{a} \cdot \overline{c}=|\overline{c}|$,we have $|\overline{c}|^2+9-2|\overline{c}|=8$.
This simplifies to $|\overline{c}|^2-2|\overline{c}|+1=0$,which is $(|\overline{c}|-1)^2=0$,so $|\overline{c}|=1$.
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 $|\overline{a} \times \overline{b}| = \sqrt{2^2+(-2)^2+1^2} = \sqrt{9} = 3$.
Finally,$|(\overline{a} \times \overline{b}) \times \overline{c}| = |\overline{a} \times \overline{b}| |\overline{c}| \sin\left(\frac{2\pi}{3}\right) = 3 \times 1 \times \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2}$.

(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) Let $\vec{a} = \hat{i}+\hat{j}+\hat{k}$.
Let $\vec{b} = 2 \hat{i}+4 \hat{j}-5 \hat{k}$ and $\vec{c} = \lambda \hat{i}+2 \hat{j}+3 \hat{k}$.
The sum of the vectors is $\vec{v} = \vec{b} + \vec{c} = (2+\lambda) \hat{i}+6 \hat{j}-2 \hat{k}$.
The unit vector along $\vec{v}$ is $\hat{v} = \frac{(2+\lambda) \hat{i}+6 \hat{j}-2 \hat{k}}{\sqrt{(2+\lambda)^2+6^2+(-2)^2}} = \frac{(2+\lambda) \hat{i}+6 \hat{j}-2 \hat{k}}{\sqrt{\lambda^2+4 \lambda+44}}$.
The scalar product of $\vec{a}$ and $\hat{v}$ is $1$:
$\vec{a} \cdot \hat{v} = 1$
$\frac{(1)(2+\lambda) + (1)(6) + (1)(-2)}{\sqrt{\lambda^2+4 \lambda+44}} = 1$
$\frac{\lambda+6}{\sqrt{\lambda^2+4 \lambda+44}} = 1$
$\lambda+6 = \sqrt{\lambda^2+4 \lambda+44}$
Squaring both sides:
$(\lambda+6)^2 = \lambda^2+4 \lambda+44$
$\lambda^2+12 \lambda+36 = \lambda^2+4 \lambda+44$
$8 \lambda = 8$
$\lambda = 1$.

(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) Let $\bar{s} = \bar{b} + \bar{c} = (2+\lambda) \hat{i} - 2 \hat{j} + 2 \hat{k}$.
The unit vector along $\bar{s}$ is $\hat{s} = \frac{\bar{s}}{|\bar{s}|} = \frac{(2+\lambda) \hat{i} - 2 \hat{j} + 2 \hat{k}}{\sqrt{(2+\lambda)^2 + (-2)^2 + 2^2}} = \frac{(2+\lambda) \hat{i} - 2 \hat{j} + 2 \hat{k}}{\sqrt{\lambda^2 + 4\lambda + 12}}$.
Given that $\bar{a} \cdot \hat{s} = 1$,we have:
$(\hat{i} + 2 \hat{j} + \hat{k}) \cdot \left( \frac{(2+\lambda) \hat{i} - 2 \hat{j} + 2 \hat{k}}{\sqrt{\lambda^2 + 4\lambda + 12}} \right) = 1$.
Taking the dot product in the numerator:
$\frac{(2+\lambda) - 4 + 2}{\sqrt{\lambda^2 + 4\lambda + 12}} = 1$.
$\frac{\lambda}{\sqrt{\lambda^2 + 4\lambda + 12}} = 1$.
Squaring both sides:
$\lambda^2 = \lambda^2 + 4\lambda + 12$.
$4\lambda = -12$.
$\lambda = -3$.

(A) Using the vector triple product formula,$\bar{a} \times(\bar{b} \times \bar{c}) = (\bar{a} \cdot \bar{c}) \bar{b} - (\bar{a} \cdot \bar{b}) \bar{c}$.
Given $\bar{a} \times(\bar{b} \times \bar{c}) = \frac{\bar{b}+\bar{c}}{\sqrt{2}}$,we have:
$(\bar{a} \cdot \bar{c}) \bar{b} - (\bar{a} \cdot \bar{b}) \bar{c} = \frac{1}{\sqrt{2}} \bar{b} + \frac{1}{\sqrt{2}} \bar{c}$.
Rearranging the terms,we get:
$(\bar{a} \cdot \bar{c} - \frac{1}{\sqrt{2}}) \bar{b} - (\bar{a} \cdot \bar{b} + \frac{1}{\sqrt{2}}) \bar{c} = 0$.
Since $\bar{b}$ and $\bar{c}$ are non-coplanar (and thus linearly independent),the coefficients must be zero:
$\bar{a} \cdot \bar{b} + \frac{1}{\sqrt{2}} = 0 \Rightarrow \bar{a} \cdot \bar{b} = -\frac{1}{\sqrt{2}}$.
Since $\bar{a}$ and $\bar{b}$ are unit vectors,$|\bar{a}| = 1$ and $|\bar{b}| = 1$.
Thus,$|\bar{a}| |\bar{b}| \cos \theta = -\frac{1}{\sqrt{2}}$,where $\theta$ is the angle between $\bar{a}$ and $\bar{b}$.
$\cos \theta = -\frac{1}{\sqrt{2}}$.
Therefore,$\theta = \frac{3 \pi}{4}$.

(A) Using the vector triple product identity,we have $\overline{a} \times (\overline{b} \times \overline{c}) = (\overline{a} \cdot \overline{c}) \overline{b} - (\overline{a} \cdot \overline{b}) \overline{c}$.
Given $\overline{a} \times (\overline{b} \times \overline{c}) = \frac{\sqrt{3}}{2} \overline{b} + \frac{\sqrt{3}}{2} \overline{c}$,we compare the coefficients of $\overline{b}$ and $\overline{c}$ since $\overline{b}$ and $\overline{c}$ are not parallel.
Thus,$\overline{a} \cdot \overline{c} = \frac{\sqrt{3}}{2}$ and $-\overline{a} \cdot \overline{b} = \frac{\sqrt{3}}{2}$,which implies $\overline{a} \cdot \overline{b} = -\frac{\sqrt{3}}{2}$.
Since $\overline{a}$ and $\overline{b}$ are unit vectors,$\overline{a} \cdot \overline{b} = |\overline{a}| |\overline{b}| \cos \theta = (1)(1) \cos \theta = \cos \theta$.
Therefore,$\cos \theta = -\frac{\sqrt{3}}{2}$.
Since $\theta \in [0, \pi]$,we have $\theta = \frac{5\pi}{6}$.

(A) Given $\overline{a}=2 \hat{i}+\hat{j}-2 \hat{k}$,so $|\overline{a}|^2 = 2^2 + 1^2 + (-2)^2 = 4+1+4 = 9$,which implies $|\overline{a}|=3$.
Given $|\overline{c}-\overline{a}|=2 \sqrt{2}$,squaring both sides gives $|\overline{c}-\overline{a}|^2 = (2 \sqrt{2})^2 = 8$.
Expanding this,we get $|\overline{c}|^2 + |\overline{a}|^2 - 2(\overline{a} \cdot \overline{c}) = 8$.
Substituting $\overline{a} \cdot \overline{c} = |\overline{c}|$ and $|\overline{a}|^2 = 9$,we have $|\overline{c}|^2 + 9 - 2|\overline{c}| = 8$,which simplifies to $|\overline{c}|^2 - 2|\overline{c}| + 1 = 0$.
This is $(|\overline{c}|-1)^2 = 0$,so $|\overline{c}| = 1$.
Next,calculate $\overline{a} \times \overline{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 1 & 1 & 0 \end{vmatrix} = \hat{i}(0 - (-2)) - \hat{j}(0 - (-2)) + \hat{k}(2 - 1) = 2 \hat{i} - 2 \hat{j} + \hat{k}$.
Thus,$|\overline{a} \times \overline{b}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4+4+1} = \sqrt{9} = 3$.
Finally,$|(\overline{a} \times \overline{b}) \times \overline{c}| = |\overline{a} \times \overline{b}| |\overline{c}| \sin(\frac{\pi}{6}) = 3 \times 1 \times \frac{1}{2} = \frac{3}{2}$.

(D) Given $\bar{a}=2 \hat{i}+\hat{j}-2 \hat{k}$ and $\bar{b}=\hat{i}+\hat{j}$.
First,calculate the magnitude of $\bar{a}$:
$|\bar{a}|=\sqrt{2^2+1^2+(-2)^2}=\sqrt{4+1+4}=3$.
Next,calculate the cross product $\bar{a} \times \bar{b}$:
$\bar{a} \times \bar{b}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 1 & 1 & 0 \end{vmatrix} = \hat{i}(0 - (-2)) - \hat{j}(0 - (-2)) + \hat{k}(2 - 1) = 2 \hat{i}-2 \hat{j}+\hat{k}$.
The magnitude is $|\bar{a} \times \bar{b}|=\sqrt{2^2+(-2)^2+1^2}=\sqrt{4+4+1}=3$.
Given $|(\bar{a} \times \bar{b}) \times \bar{c}|=3$ and the angle $\theta$ between $\bar{c}$ and $\bar{a} \times \bar{b}$ is $\frac{\pi}{6}$.
Using the formula $|\bar{u} \times \bar{v}| = |\bar{u}||\bar{v}| \sin \theta$:
$|(\bar{a} \times \bar{b}) \times \bar{c}| = |\bar{a} \times \bar{b}| |\bar{c}| \sin \frac{\pi}{6} = 3$.
$3 \cdot |\bar{c}| \cdot \frac{1}{2} = 3 \Rightarrow |\bar{c}|=2$.
Now,use the given $|\bar{c}-\bar{a}|=4$:
$|\bar{c}-\bar{a}|^2 = |\bar{c}|^2 + |\bar{a}|^2 - 2(\bar{a} \cdot \bar{c}) = 4^2$.
$2^2 + 3^2 - 2(\bar{a} \cdot \bar{c}) = 16$.
$4 + 9 - 2(\bar{a} \cdot \bar{c}) = 16$.
$13 - 2(\bar{a} \cdot \bar{c}) = 16$.
$-2(\bar{a} \cdot \bar{c}) = 3$.
$\bar{a} \cdot \bar{c} = -\frac{3}{2}$.

(D) Given that $(\overline{a}-\overline{d}) \cdot(\overline{b}-\overline{c})=0$ and $(\overline{b}-\overline{d}) \cdot(\overline{c}-\overline{a})=0$.
This can be written in terms of vectors as $\overline{AD} \cdot \overline{BC} = 0$ and $\overline{BD} \cdot \overline{CA} = 0$.
This implies that $\overline{AD} \perp \overline{BC}$ and $\overline{BD} \perp \overline{CA}$.
Since $D$ is the point of intersection of the altitudes $AD$ and $BD$ of $\triangle ABC$,$D$ is the orthocentre of $\triangle ABC$.

(C) Let the given expression be $E = \tan \left[\sin ^{-1}\left\{\frac{x}{\sqrt{2}}+\frac{\sqrt{1-x^{2}}}{\sqrt{2}}\right\}-\sin ^{-1} x\right]$.
Substitute $x = \sin \theta$,where $\theta = \sin^{-1} x$.
Since $0 \leq x \leq \frac{1}{2}$,we have $0 \leq \theta \leq \frac{\pi}{6}$.
The expression inside the $\sin^{-1}$ becomes $\frac{\sin \theta}{\sqrt{2}} + \frac{\cos \theta}{\sqrt{2}} = \sin \theta \cos \frac{\pi}{4} + \cos \theta \sin \frac{\pi}{4} = \sin \left(\theta + \frac{\pi}{4}\right)$.
Now,the expression becomes $E = \tan \left[\sin ^{-1}\left\{\sin \left(\theta + \frac{\pi}{4}\right)\right\} - \theta\right]$.
Since $0 \leq \theta \leq \frac{\pi}{6}$,then $\frac{\pi}{4} \leq \theta + \frac{\pi}{4} \leq \frac{\pi}{6} + \frac{\pi}{4} = \frac{5\pi}{12}$.
Since $\frac{\pi}{4} \leq \theta + \frac{\pi}{4} \leq \frac{5\pi}{12}$,the value lies within the principal range of $\sin^{-1}$,so $\sin^{-1}(\sin(\theta + \frac{\pi}{4})) = \theta + \frac{\pi}{4}$.
Thus,$E = \tan \left(\theta + \frac{\pi}{4} - \theta\right) = \tan \frac{\pi}{4} = 1$.