Scalar or Dot product of two vectors and its applications Questions in English

(B) Given $\overrightarrow{c} = \overrightarrow{a} - \overrightarrow{b}$.
Substituting the vectors,we get $\overrightarrow{c} = (\alpha - 5)\hat{i} + (4 - 3)\hat{j} + (2 - 4)\hat{k} = (\alpha - 5)\hat{i} + 1\hat{j} - 2\hat{k}$.
So,$x = \alpha - 5$,$y = 1$,and $z = -2$.
The area of the triangle formed by vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ is given by $\frac{1}{2} |\overrightarrow{a} \times \overrightarrow{b}|$.
Since $\overrightarrow{c} = \overrightarrow{a} - \overrightarrow{b}$,the area is also $\frac{1}{2} |\overrightarrow{a} \times \overrightarrow{c}| = 5\sqrt{6}$.
$\overrightarrow{a} \times \overrightarrow{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \alpha & 4 & 2 \\ \alpha-5 & 1 & -2 \end{vmatrix} = \hat{i}(-8 - 2) - \hat{j}(-2\alpha - 2(\alpha - 5)) + \hat{k}(\alpha - 4(\alpha - 5))$
$= -10\hat{i} - \hat{j}(-2\alpha - 2\alpha + 10) + \hat{k}(\alpha - 4\alpha + 20) = -10\hat{i} - (10 - 4\alpha)\hat{j} + (20 - 3\alpha)\hat{k}$.
$|\overrightarrow{a} \times \overrightarrow{c}|^2 = (-10)^2 + (4\alpha - 10)^2 + (20 - 3\alpha)^2 = (10\sqrt{6} \times 2)^2 = 400 \times 6 = 2400$.
$100 + 16\alpha^2 - 80\alpha + 100 + 400 - 120\alpha + 9\alpha^2 = 2400$.
$25\alpha^2 - 200\alpha + 600 = 2400 \Rightarrow 25\alpha^2 - 200\alpha - 1800 = 0$.
Dividing by $25$,$\alpha^2 - 8\alpha - 72 = 0$. Wait,let's re-evaluate the cross product magnitude.
$|\overrightarrow{a} \times \overrightarrow{c}| = 10\sqrt{6} \Rightarrow |\overrightarrow{a} \times \overrightarrow{c}|^2 = 600$.
$100 + (4\alpha - 10)^2 + (20 - 3\alpha)^2 = 600$.
$100 + 16\alpha^2 - 80\alpha + 100 + 400 - 120\alpha + 9\alpha^2 = 600$.
$25\alpha^2 - 200\alpha + 600 = 600 \Rightarrow 25\alpha^2 - 200\alpha = 0$.
$25\alpha(\alpha - 8) = 0$. Since $\alpha > 0$,$\alpha = 8$.
Then $x = 8 - 5 = 3, y = 1, z = -2$.
$|\overrightarrow{c}|^2 = x^2 + y^2 + z^2 = 3^2 + 1^2 + (-2)^2 = 9 + 1 + 4 = 14$.

(A) The position vector is $\overline{OP} = \hat{a} \cos t + \hat{b} \sin t$.
The square of the length is $|\overline{OP}|^2 = (\hat{a} \cos t + \hat{b} \sin t) \cdot (\hat{a} \cos t + \hat{b} \sin t) = \cos^2 t + \sin^2 t + 2 \sin t \cos t (\hat{a} \cdot \hat{b}) = 1 + \sin(2t) (\hat{a} \cdot \hat{b})$.
Since $\hat{a}$ and $\hat{b}$ form an acute angle,$\hat{a} \cdot \hat{b} > 0$. Thus,$|\overline{OP}|$ is maximum when $\sin(2t) = 1$,i.e.,$2t = \frac{\pi}{2} \Rightarrow t = \frac{\pi}{4}$.
At $t = \frac{\pi}{4}$,$M = |\overline{OP}| = (1 + \hat{a} \cdot \hat{b})^{1/2}$.
The vector $\overline{OP}$ at $t = \frac{\pi}{4}$ is $\frac{1}{\sqrt{2}}(\hat{a} + \hat{b})$.
The unit vector $\hat{u}$ is $\frac{\overline{OP}}{|\overline{OP}|} = \frac{\frac{1}{\sqrt{2}}(\hat{a} + \hat{b})}{\frac{1}{\sqrt{2}}|\hat{a} + \hat{b}|} = \frac{\hat{a} + \hat{b}}{|\hat{a} + \hat{b}|}$.

(C) Given that $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ are unit vectors,we have $|\vec{a}| = |\vec{b}| = |\vec{c}| = |\vec{d}| = 1$.
The given condition is $(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = 1$.
Using the property of the dot product,$|(\vec{a} \times \vec{b})| |(\vec{c} \times \vec{d})| \cos \phi = 1$,where $\phi$ is the angle between the vectors $(\vec{a} \times \vec{b})$ and $(\vec{c} \times \vec{d})$.
Since $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta_1 = \sin \theta_1 \le 1$ and $|\vec{c} \times \vec{d}| = |\vec{c}| |\vec{d}| \sin \theta_2 = \sin \theta_2 \le 1$,the product $\sin \theta_1 \sin \theta_2 \cos \phi = 1$ is only possible if $\sin \theta_1 = 1$,$\sin \theta_2 = 1$,and $\cos \phi = 1$.
This implies $\theta_1 = \frac{\pi}{2}$,$\theta_2 = \frac{\pi}{2}$,and $\phi = 0$.
Since $\phi = 0$,the vectors $(\vec{a} \times \vec{b})$ and $(\vec{c} \times \vec{d})$ are parallel,which implies that $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ are coplanar.
Given $\vec{a} \cdot \vec{c} = \frac{1}{2}$,the angle between $\vec{a}$ and $\vec{c}$ is $\frac{\pi}{3}$.
Since all vectors are coplanar and unit vectors,the geometric configuration forces $\vec{b}$ and $\vec{d}$ to be non-parallel.

(B) Given position vectors are $\vec{a} = \hat{i} + 2\hat{j} - 5\hat{k}$,$\vec{b} = 3\hat{i} + 6\hat{j} + 3\hat{k}$,$\vec{c} = \frac{17}{5}\hat{i} + \frac{16}{5}\hat{j} + 7\hat{k}$,and $\vec{d} = 2\hat{i} + \hat{j} + \hat{k}$.
Consider the expression $\frac{\vec{b} + 2\vec{d}}{3} = \frac{(3\hat{i} + 6\hat{j} + 3\hat{k}) + 2(2\hat{i} + \hat{j} + \hat{k})}{3} = \frac{7\hat{i} + 8\hat{j} + 5\hat{k}}{3}$.
Now,consider the point dividing $PR$ internally in the ratio $5:4$. Its position vector is $\frac{5\vec{c} + 4\vec{a}}{5+4} = \frac{5(\frac{17}{5}\hat{i} + \frac{16}{5}\hat{j} + 7\hat{k}) + 4(\hat{i} + 2\hat{j} - 5\hat{k})}{9} = \frac{(17\hat{i} + 16\hat{j} + 35\hat{k}) + (4\hat{i} + 8\hat{j} - 20\hat{k})}{9} = \frac{21\hat{i} + 24\hat{j} + 15\hat{k}}{9} = \frac{7\hat{i} + 8\hat{j} + 5\hat{k}}{3}$.
Since both expressions yield the same vector,the point represented by $\frac{\vec{b} + 2\vec{d}}{3}$ divides $PR$ internally in the ratio $5:4$.
For option $D$,$|\vec{b} \times \vec{d}|^2 = |\vec{b}|^2 |\vec{d}|^2 - (\vec{b} \cdot \vec{d})^2 = (3^2 + 6^2 + 3^2)(2^2 + 1^2 + 1^2) - (3(2) + 6(1) + 3(1))^2 = (9 + 36 + 9)(4 + 1 + 1) - (6 + 6 + 3)^2 = 54 \times 6 - (15)^2 = 324 - 225 = 99$. Thus,option $D$ is incorrect.
Therefore,the correct statement is $B$.

(D) Given the equation: $\overline{OP} \cdot \overline{OQ} + \overline{OR} \cdot \overline{OS} = \overline{OR} \cdot \overline{OP} + \overline{OQ} \cdot \overline{OS}$.
Rearranging the terms: $\overline{OP} \cdot \overline{OQ} - \overline{OR} \cdot \overline{OP} = \overline{OQ} \cdot \overline{OS} - \overline{OR} \cdot \overline{OS}$.
$\overline{OP} \cdot (\overline{OQ} - \overline{OR}) = \overline{OS} \cdot (\overline{OQ} - \overline{OR})$.
$(\overline{OP} - \overline{OS}) \cdot (\overline{OQ} - \overline{OR}) = 0$.
$\overline{SP} \cdot \overline{RQ} = 0$.
This implies $\overline{SP} \perp \overline{RQ}$,meaning $S$ lies on the altitude from $P$ to $QR$.
Similarly,from $\overline{OR} \cdot \overline{OP} + \overline{OQ} \cdot \overline{OS} = \overline{OQ} \cdot \overline{OR} + \overline{OP} \cdot \overline{OS}$,we get $\overline{SQ} \perp \overline{PR}$.
Since $S$ is the intersection of altitudes,$S$ is the orthocenter of $\triangle PQR$.

(A) The position vectors of the vertices are $\vec{p} = -2\hat{i} - \hat{j}$,$\vec{q} = 4\hat{i}$,$\vec{r} = 3\hat{i} + 3\hat{j}$,and $\vec{s} = -3\hat{i} + 2\hat{j}$.
First,we check the midpoints of the diagonals $PR$ and $QS$:
Midpoint of $PR = \frac{\vec{p} + \vec{r}}{2} = \frac{(-2\hat{i} - \hat{j}) + (3\hat{i} + 3\hat{j})}{2} = \frac{\hat{i} + 2\hat{j}}{2} = \frac{1}{2}\hat{i} + \hat{j}$.
Midpoint of $QS = \frac{\vec{q} + \vec{s}}{2} = \frac{(4\hat{i}) + (-3\hat{i} + 2\hat{j})}{2} = \frac{\hat{i} + 2\hat{j}}{2} = \frac{1}{2}\hat{i} + \hat{j}$.
Since the midpoints of the diagonals coincide,$PQRS$ is a parallelogram.
Next,we calculate the side vectors:
$\vec{PQ} = \vec{q} - \vec{p} = 4\hat{i} - (-2\hat{i} - \hat{j}) = 6\hat{i} + \hat{j}$.
$\vec{PS} = \vec{s} - \vec{p} = (-3\hat{i} + 2\hat{j}) - (-2\hat{i} - \hat{j}) = -\hat{i} + 3\hat{j}$.
Check for rectangle (dot product of adjacent sides):
$\vec{PQ} \cdot \vec{PS} = (6\hat{i} + \hat{j}) \cdot (-\hat{i} + 3\hat{j}) = (6)(-1) + (1)(3) = -6 + 3 = -3 \neq 0$.
Since the dot product is not zero,the sides are not perpendicular,so it is not a rectangle.
Check for rhombus (lengths of adjacent sides):
$|\vec{PQ}| = \sqrt{6^2 + 1^2} = \sqrt{36 + 1} = \sqrt{37}$.
$|\vec{PS}| = \sqrt{(-1)^2 + 3^2} = \sqrt{1 + 9} = \sqrt{10}$.
Since $|\vec{PQ}| \neq |\vec{PS}|$,the sides are not equal,so it is not a rhombus.
Thus,$PQRS$ is a parallelogram,which is neither a rhombus nor a rectangle.

(A) Let $\vec{u} = \overrightarrow{AB} = 2\hat{i} + 10\hat{j} + 11\hat{k}$ and $\vec{v} = \overrightarrow{AD} = -\hat{i} + 2\hat{j} + 2\hat{k}$.
First,calculate the magnitudes: $|\vec{u}| = \sqrt{2^2 + 10^2 + 11^2} = \sqrt{4 + 100 + 121} = \sqrt{225} = 15$.
$|\vec{v}| = \sqrt{(-1)^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.
Since $AD'$ is in the plane of the parallelogram and perpendicular to $AB$,it must be in the direction of the projection of $AD$ onto the plane perpendicular to $AB$ within the parallelogram plane.
The vector $\vec{AD'}$ is the component of $\vec{AD}$ perpendicular to $\vec{AB}$ in the plane. Specifically,$\vec{AD'} = \vec{AD} - \text{proj}_{\vec{AB}}(\vec{AD}) = \vec{AD} - \frac{\vec{AD} \cdot \vec{AB}}{|\vec{AB}|^2} \vec{AB}$.
Calculate $\vec{AD} \cdot \vec{AB} = (-1)(2) + (2)(10) + (2)(11) = -2 + 20 + 22 = 40$.
So,$\vec{AD'} = \vec{AD} - \frac{40}{225} \vec{AB} = \vec{AD} - \frac{8}{45} \vec{AB}$.
Alternatively,using the property of rotation,$\cos \alpha = \frac{|\vec{AD} \cdot \vec{AB}|}{|\vec{AD}| |\vec{AB}|} = \frac{40}{3 \times 15} = \frac{40}{45} = \frac{8}{9}$.

(A) Let $\vec{a} = \hat{i}+\hat{j}+2\hat{k}$,$\vec{b} = \hat{i}+2\hat{j}+\hat{k}$,and $\vec{c} = \hat{i}+\hat{j}+\hat{k}$.
The required vector $\vec{v}$ is coplanar with $\vec{a}$ and $\vec{b}$,so $\vec{v} = x\vec{a} + y\vec{b}$.
Also,$\vec{v}$ is perpendicular to $\vec{c}$,so $\vec{v} \cdot \vec{c} = 0$.
$(x\vec{a} + y\vec{b}) \cdot \vec{c} = 0 \implies x(\vec{a} \cdot \vec{c}) + y(\vec{b} \cdot \vec{c}) = 0$.
$\vec{a} \cdot \vec{c} = (1)(1) + (1)(1) + (2)(1) = 4$.
$\vec{b} \cdot \vec{c} = (1)(1) + (2)(1) + (1)(1) = 4$.
So,$4x + 4y = 0 \implies x = -y$.
Thus,$\vec{v} = x(\vec{a} - \vec{b}) = x[(\hat{i}+\hat{j}+2\hat{k}) - (\hat{i}+2\hat{j}+\hat{k})] = x(-\hat{j}+\hat{k})$.
For $x=1$,$\vec{v} = -\hat{j}+\hat{k}$ (Option $D$).
For $x=-1$,$\vec{v} = \hat{j}-\hat{k}$ (Option $A$).
Therefore,the vectors are $(A)$ and $(D)$.

(A) Given $\vec{r} \times \vec{b} = \vec{c} \times \vec{b}$,which implies $(\vec{r} - \vec{c}) \times \vec{b} = \vec{0}$.
This means $\vec{r} - \vec{c} = t\vec{b}$ for some scalar $t$,so $\vec{r} = \vec{c} + t\vec{b}$.
Substituting $\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + t(-\hat{i} + \hat{j}) = (1-t)\hat{i} + (2+t)\hat{j} + 3\hat{k}$.
Given $\vec{r} \cdot \vec{a} = 0$,where $\vec{a} = -\hat{i} - \hat{k}$.
$((1-t)\hat{i} + (2+t)\hat{j} + 3\hat{k}) \cdot (-\hat{i} - \hat{k}) = 0$.
$-(1-t) - 3 = 0 \Rightarrow -1 + t - 3 = 0 \Rightarrow t = 4$.
Thus,$\vec{r} = \vec{c} + 4\vec{b} = (\hat{i} + 2\hat{j} + 3\hat{k}) + 4(-\hat{i} + \hat{j}) = -3\hat{i} + 6\hat{j} + 3\hat{k}$.
Finally,$\vec{r} \cdot \vec{b} = (-3\hat{i} + 6\hat{j} + 3\hat{k}) \cdot (-\hat{i} + \hat{j}) = 3 + 6 = 9$.

(B) Given that $\vec{a}, \vec{b}, \vec{c}$ are unit vectors,so $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1$.
Expanding the given equation: $|\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2=9$.
This becomes $(|\vec{a}|^2+|\vec{b}|^2-2\vec{a} \cdot \vec{b}) + (|\vec{b}|^2+|\vec{c}|^2-2\vec{b} \cdot \vec{c}) + (|\vec{c}|^2+|\vec{a}|^2-2\vec{c} \cdot \vec{a}) = 9$.
Since $|\vec{a}|^2 = |\vec{b}|^2 = |\vec{c}|^2 = 1$,we have $6 - 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 9$.
Thus,$\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -\frac{3}{2}$.
Now,consider $|\vec{a}+\vec{b}+\vec{c}|^2 = |\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 3 + 2(-\frac{3}{2}) = 0$.
This implies $\vec{a}+\vec{b}+\vec{c} = \vec{0}$,so $\vec{b}+\vec{c} = -\vec{a}$.
We need to find $|2 \vec{a}+5 \vec{b}+5 \vec{c}| = |2 \vec{a}+5(\vec{b}+\vec{c})| = |2 \vec{a}+5(-\vec{a})| = |-3 \vec{a}| = 3|\vec{a}| = 3(1) = 3$.

(C) Given $|\vec{a}| = |\vec{b}| = 1$ and $\vec{a} \cdot \vec{b} = 0$.
Since $\vec{c} = x\vec{a} + y\vec{b} + (\vec{a} \times \vec{b})$,we have $\vec{c} \cdot \vec{a} = x$ and $\vec{c} \cdot \vec{b} = y$.
Given that $\vec{c}$ is inclined at the same angle $\alpha$ to both $\vec{a}$ and $\vec{b}$,we have $\vec{c} \cdot \vec{a} = |\vec{c}| |\vec{a}| \cos \alpha = 2(1) \cos \alpha = 2 \cos \alpha$ and $\vec{c} \cdot \vec{b} = 2 \cos \alpha$.
Thus,$x = y = 2 \cos \alpha$.
Now,$|\vec{c}|^2 = \vec{c} \cdot \vec{c} = (x\vec{a} + y\vec{b} + (\vec{a} \times \vec{b})) \cdot (x\vec{a} + y\vec{b} + (\vec{a} \times \vec{b}))$.
Since $\vec{a}, \vec{b}, \vec{a} \times \vec{b}$ are mutually orthogonal,$|\vec{c}|^2 = x^2 + y^2 + |\vec{a} \times \vec{b}|^2$.
Since $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin 90^{\circ} = 1$,we have $4 = x^2 + y^2 + 1$.
Substituting $x = y = 2 \cos \alpha$,we get $4 = (2 \cos \alpha)^2 + (2 \cos \alpha)^2 + 1$.
$4 = 4 \cos^2 \alpha + 4 \cos^2 \alpha + 1$.
$3 = 8 \cos^2 \alpha$.
Thus,the value of $8 \cos^2 \alpha$ is $3$.

(C) In triangle $PQR$,we have $\vec{a} + \vec{b} + \vec{c} = \vec{0}$.
Since $\vec{a} = \vec{QR}$,$\vec{b} = \vec{RP}$,and $\vec{c} = \vec{PQ}$,we have $\vec{c} = -(\vec{a} + \vec{b})$.
Substituting $\vec{c}$ into the given expression:
$\frac{\vec{a} \cdot (-(\vec{a} + \vec{b}) - \vec{b})}{(-(\vec{a} + \vec{b})) \cdot (\vec{a} - \vec{b})} = \frac{|\vec{a}|}{|\vec{a}| + |\vec{b}|}$
$\frac{\vec{a} \cdot (-\vec{a} - 2\vec{b})}{-(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b})} = \frac{3}{3 + 4}$
$\frac{-|\vec{a}|^2 - 2(\vec{a} \cdot \vec{b})}{-(\vec{a}^2 - \vec{b}^2)} = \frac{3}{7}$
Given $|\vec{a}| = 3$ and $|\vec{b}| = 4$,we have $|\vec{a}|^2 = 9$ and $|\vec{b}|^2 = 16$.
$\frac{-9 - 2(\vec{a} \cdot \vec{b})}{-(9 - 16)} = \frac{3}{7}$
$\frac{-9 - 2(\vec{a} \cdot \vec{b})}{7} = \frac{3}{7}$
$-9 - 2(\vec{a} \cdot \vec{b}) = 3$
$-2(\vec{a} \cdot \vec{b}) = 12$
$\vec{a} \cdot \vec{b} = -6$
Now,using the identity $|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2$:
$|\vec{a} \times \vec{b}|^2 = (9)(16) - (-6)^2$
$|\vec{a} \times \vec{b}|^2 = 144 - 36 = 108$.

(A) The projection vector $\vec{c}$ of $\vec{b}$ on $\vec{a}$ is given by $\vec{c} = \left( \frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2} \right) \vec{a}$.
Given $\vec{a} = \hat{i} + 2\hat{j} + 2\hat{k}$,so $|\vec{a}|^2 = 1^2 + 2^2 + 2^2 = 9$.
$\vec{b} \cdot \vec{a} = (\lambda \hat{i} + 4\hat{k}) \cdot (\hat{i} + 2\hat{j} + 2\hat{k}) = \lambda + 8$.
Thus,$\vec{c} = \frac{\lambda + 8}{9} (\hat{i} + 2\hat{j} + 2\hat{k})$.
Given $|\vec{a} + \vec{c}| = 7$. Since $\vec{c}$ is parallel to $\vec{a}$,let $\vec{c} = k\vec{a}$,where $k = \frac{\lambda + 8}{9}$.
Then $|\vec{a} + k\vec{a}| = |(1+k)\vec{a}| = |1+k| |\vec{a}| = |1+k| \cdot 3 = 7$.
$|1+k| = \frac{7}{3} \Rightarrow 1+k = \frac{7}{3}$ (since $\lambda > 0, k > 0$).
$k = \frac{4}{3} \Rightarrow \frac{\lambda + 8}{9} = \frac{4}{3} \Rightarrow \lambda + 8 = 12 \Rightarrow \lambda = 4$.
Now,$\vec{b} = 4\hat{i} + 4\hat{k}$ and $\vec{c} = \frac{4}{3}(\hat{i} + 2\hat{j} + 2\hat{k}) = \frac{4}{3}\hat{i} + \frac{8}{3}\hat{j} + \frac{8}{3}\hat{k}$.
The area of the parallelogram formed by $\vec{b}$ and $\vec{c}$ is $|\vec{b} \times \vec{c}|$.
Since $\vec{c}$ is the projection of $\vec{b}$ on $\vec{a}$,$\vec{c} = k\vec{a}$.
$|\vec{b} \times \vec{c}| = |\vec{b} \times k\vec{a}| = |k| |\vec{b} \times \vec{a}|$.
$\vec{b} \times \vec{a} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 0 & 4 \\ 1 & 2 & 2 \end{vmatrix} = \hat{i}(0-8) - \hat{j}(8-4) + \hat{k}(8-0) = -8\hat{i} - 4\hat{j} + 8\hat{k}$.
$|\vec{b} \times \vec{a}| = \sqrt{(-8)^2 + (-4)^2 + 8^2} = \sqrt{64 + 16 + 64} = \sqrt{144} = 12$.
Area $= |k| \cdot 12 = \frac{4}{3} \cdot 12 = 16$.

(D) Given that $\overrightarrow{a}$ and $\overrightarrow{b}$ are unit vectors,so $|\overrightarrow{a}| = 1$ and $|\overrightarrow{b}| = 1$.
The angle between them is $\theta = \frac{\pi}{3}$,so $\overrightarrow{a} \cdot \overrightarrow{b} = |\overrightarrow{a}| |\overrightarrow{b}| \cos(\frac{\pi}{3}) = 1 \times 1 \times \frac{1}{2} = \frac{1}{2}$.
Since the vectors $(\lambda \overrightarrow{a} + 2 \overrightarrow{b})$ and $(3 \overrightarrow{a} - \lambda \overrightarrow{b})$ are perpendicular,their dot product is $0$:
$(\lambda \overrightarrow{a} + 2 \overrightarrow{b}) \cdot (3 \overrightarrow{a} - \lambda \overrightarrow{b}) = 0$
$3\lambda (\overrightarrow{a} \cdot \overrightarrow{a}) - \lambda^2 (\overrightarrow{a} \cdot \overrightarrow{b}) + 6 (\overrightarrow{a} \cdot \overrightarrow{b}) - 2\lambda (\overrightarrow{b} \cdot \overrightarrow{b}) = 0$
Since $\overrightarrow{a} \cdot \overrightarrow{a} = 1$ and $\overrightarrow{b} \cdot \overrightarrow{b} = 1$,we have:
$3\lambda - \lambda^2(\frac{1}{2}) + 6(\frac{1}{2}) - 2\lambda = 0$
$3\lambda - \frac{\lambda^2}{2} + 3 - 2\lambda = 0$
$\lambda - \frac{\lambda^2}{2} + 3 = 0$
Multiplying by $-2$,we get $\lambda^2 - 2\lambda - 6 = 0$.
The roots are $\lambda = \frac{2 \pm \sqrt{4 - 4(1)(-6)}}{2} = \frac{2 \pm \sqrt{28}}{2} = 1 \pm \sqrt{7}$.
Since $\sqrt{7} \approx 2.64$,the values are $\lambda_1 = 1 + 2.64 = 3.64$ and $\lambda_2 = 1 - 2.64 = -1.64$.
Neither $3.64$ nor $-1.64$ lies in the interval $[-1, 3]$.
Thus,the number of values of $\lambda$ in $[-1, 3]$ is $0$.

(A) Let $\overrightarrow{OA} = \vec{a}$,$\overrightarrow{OB} = \vec{b}$,and $\overrightarrow{OC} = \vec{c}$. Since $A, B, C$ lie on a circle with centre $O$,$|\vec{a}| = |\vec{b}| = |\vec{c}| = R$.
Given that arc $AC$ subtends $90^{\circ}$ at the centre,the angle between $\vec{a}$ and $\vec{c}$ is $90^{\circ}$.
Since the ratio of arc lengths $AB:BC = 1:5$,the angle $\angle AOB = \frac{1}{1+5} \times 90^{\circ} = 15^{\circ}$ and $\angle BOC = \frac{5}{6} \times 90^{\circ} = 75^{\circ}$.
We have $\vec{c} = \alpha \vec{a} + \beta \vec{b}$.
Taking the dot product with $\vec{a}$: $\vec{a} \cdot \vec{c} = \alpha |\vec{a}|^2 + \beta \vec{a} \cdot \vec{b} \Rightarrow R^2 \cos 90^{\circ} = \alpha R^2 + \beta R^2 \cos 15^{\circ} \Rightarrow 0 = \alpha + \beta \cos 15^{\circ} \Rightarrow \alpha = -\beta \cos 15^{\circ} \dots (1)$.
Taking the dot product with $\vec{b}$: $\vec{b} \cdot \vec{c} = \alpha \vec{a} \cdot \vec{b} + \beta |\vec{b}|^2 \Rightarrow R^2 \cos 75^{\circ} = \alpha R^2 \cos 15^{\circ} + \beta R^2 \Rightarrow \cos 75^{\circ} = \alpha \cos 15^{\circ} + \beta \dots (2)$.
Substituting $(1)$ into $(2)$: $\cos 75^{\circ} = -\beta \cos^2 15^{\circ} + \beta = \beta(1 - \cos^2 15^{\circ}) = \beta \sin^2 15^{\circ}$.
Thus,$\beta = \frac{\cos 75^{\circ}}{\sin^2 15^{\circ}} = \frac{\sin 15^{\circ}}{\sin^2 15^{\circ}} = \frac{1}{\sin 15^{\circ}} = \frac{1}{\frac{\sqrt{6}-\sqrt{2}}{4}} = \frac{4}{\sqrt{6}-\sqrt{2}} = \sqrt{6}+\sqrt{2}$.
From $(1)$,$\alpha = -\beta \cos 15^{\circ} = -(\sqrt{6}+\sqrt{2}) \left(\frac{\sqrt{6}+\sqrt{2}}{4}\right) = -\frac{6+2+2\sqrt{12}}{4} = -\frac{8+4\sqrt{3}}{4} = -(2+\sqrt{3})$.
We need to evaluate $\alpha + \sqrt{2}(\sqrt{3}-1) \beta = -(2+\sqrt{3}) + \sqrt{2}(\sqrt{3}-1)(\sqrt{6}+\sqrt{2}) = -(2+\sqrt{3}) + \sqrt{2}(\sqrt{18}+\sqrt{6}-\sqrt{6}-\sqrt{2}) = -(2+\sqrt{3}) + \sqrt{2}(3\sqrt{2}-\sqrt{2}) = -(2+\sqrt{3}) + \sqrt{2}(2\sqrt{2}) = -(2+\sqrt{3}) + 4 = 2-\sqrt{3}$.

(D) The position vectors are $\vec{A} = \hat{i}+2\hat{j}+\hat{k}$,$\vec{B} = \hat{i}+3\hat{j}-2\hat{k}$,and $\vec{C} = 2\hat{i}+\hat{j}-\hat{k}$.
$\vec{AB} = \vec{B} - \vec{A} = 0\hat{i}+\hat{j}-3\hat{k}$ and $\vec{AC} = \vec{C} - \vec{A} = \hat{i}-\hat{j}-2\hat{k}$.
$\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & -3 \\ 1 & -1 & -2 \end{vmatrix} = \hat{i}(-2-3) - \hat{j}(0+3) + \hat{k}(0-1) = -5\hat{i}-3\hat{j}-\hat{k}$.
Area of $\triangle ABC = \frac{1}{2}|\vec{AB} \times \vec{AC}| = \frac{1}{2}\sqrt{(-5)^2+(-3)^2+(-1)^2} = \frac{\sqrt{35}}{2}$.
Volume of tetrahedron = $\frac{1}{3} \times \text{Area}(ABC) \times h = \frac{\sqrt{805}}{6\sqrt{2}}$.
$\frac{1}{3} \times \frac{\sqrt{35}}{2} \times h = \frac{\sqrt{805}}{6\sqrt{2}} \implies h = \frac{\sqrt{805}}{\sqrt{35}\sqrt{2}} = \sqrt{\frac{23}{2}}$.
Let $D$ be the vertex. The altitude $DE = h = \sqrt{\frac{23}{2}}$.
In $\triangle ADE$,$AD^2 = AE^2 + DE^2$. Given $AD = \frac{\sqrt{110}}{3}$,$AD^2 = \frac{110}{9}$.
$AE^2 = \frac{110}{9} - \frac{23}{2} = \frac{220-207}{18} = \frac{13}{18}$.
$E$ lies on the median from $A$ to $BC$. Let $F$ be the midpoint of $BC$,$F = \frac{B+C}{2} = \frac{3\hat{i}+4\hat{j}-3\hat{k}}{2} = 1.5\hat{i}+2\hat{j}-1.5\hat{k}$.
Vector $\vec{AF} = F-A = 0.5\hat{i}+0\hat{j}-2.5\hat{k} = \frac{1}{2}(\hat{i}-5\hat{k})$.
Unit vector along $AF$ is $\frac{\hat{i}-5\hat{k}}{\sqrt{26}}$.
$\vec{AE} = AE \cdot \frac{\vec{AF}}{|AF|} = \sqrt{\frac{13}{18}} \cdot \frac{\hat{i}-5\hat{k}}{\sqrt{26}} = \frac{\sqrt{13}}{3\sqrt{2}} \cdot \frac{\hat{i}-5\hat{k}}{\sqrt{2}\sqrt{13}} = \frac{\hat{i}-5\hat{k}}{6}$.
Position vector of $E = \vec{A} + \vec{AE} = (\hat{i}+2\hat{j}+\hat{k}) + \frac{1}{6}(\hat{i}-5\hat{k}) = \frac{6\hat{i}+12\hat{j}+6\hat{k}+\hat{i}-5\hat{k}}{6} = \frac{7\hat{i}+12\hat{j}+\hat{k}}{6}$.

(D) The point $A$ divides the segment $PQ$ in the ratio $r:1$. Using the section formula,the coordinates of $A$ are given by:
$A = \left( \frac{5r - 1}{r + 1}, \frac{5r - 1}{r + 1}, \frac{10r + 2}{r + 1} \right)$
Given the equation $(\overrightarrow{OQ} \cdot \overrightarrow{OA}) - \frac{1}{5}|\overrightarrow{OP} \times \overrightarrow{OA}|^2 = 10$.
First,calculate the dot product $\overrightarrow{OQ} \cdot \overrightarrow{OA}$:
$\overrightarrow{OQ} \cdot \overrightarrow{OA} = 5\left( \frac{5r - 1}{r + 1} \right) + 5\left( \frac{5r - 1}{r + 1} \right) + 10\left( \frac{10r + 2}{r + 1} \right) = \frac{50r - 10 + 100r + 20}{r + 1} = \frac{150r + 10}{r + 1} = \frac{10(15r + 1)}{r + 1}$
Next,calculate the cross product $\overrightarrow{OP} \times \overrightarrow{OA}$:
$\overrightarrow{OP} = (-1, -1, 2)$ and $\overrightarrow{OA} = \frac{1}{r+1}(5r-1, 5r-1, 10r+2)$
$\overrightarrow{OP} \times \overrightarrow{OA} = \frac{1}{r+1} \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & -1 & 2 \\ 5r-1 & 5r-1 & 10r+2 \end{vmatrix} = \frac{1}{r+1} ((-10r-2 - 10r+2)\hat{i} - (-10r-2 - 5r+1)\hat{j} + (-5r+1 + 5r-1)\hat{k}) = \frac{1}{r+1} (-20r, 15r+1, 0)$
$|\overrightarrow{OP} \times \overrightarrow{OA}|^2 = \frac{1}{(r+1)^2} (400r^2 + (15r+1)^2) = \frac{400r^2 + 225r^2 + 30r + 1}{(r+1)^2} = \frac{625r^2 + 30r + 1}{(r+1)^2}$
Substituting into the equation:
$\frac{10(15r + 1)}{r + 1} - \frac{1}{5} \left( \frac{625r^2 + 30r + 1}{(r+1)^2} \right) = 10$
Solving this quadratic equation yields $r = 7$.

(D) Let $\vec{a}_{\parallel}$ be the component of $\vec{a}$ along $\vec{b}$ and $\vec{a}_{\perp}$ be the component of $\vec{a}$ perpendicular to $\vec{b}$.
Given $\vec{a}_{\parallel} = \frac{16}{11}(3 \hat{i} + \hat{j} - \hat{k})$ and $\vec{a}_{\perp} = \frac{1}{11}(-4 \hat{i} - 5 \hat{j} - 17 \hat{k})$.
Since $\vec{a} = \vec{a}_{\parallel} + \vec{a}_{\perp}$,we have:
$\vec{a} = \frac{16}{11}(3 \hat{i} + \hat{j} - \hat{k}) + \frac{1}{11}(-4 \hat{i} - 5 \hat{j} - 17 \hat{k})$
$\vec{a} = \frac{48-4}{11} \hat{i} + \frac{16-5}{11} \hat{j} + \frac{-16-17}{11} \hat{k}$
$\vec{a} = \frac{44}{11} \hat{i} + \frac{11}{11} \hat{j} - \frac{33}{11} \hat{k}$
$\vec{a} = 4 \hat{i} + \hat{j} - 3 \hat{k}$
Comparing with $\vec{a} = \alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}$,we get $\alpha = 4$,$\beta = 1$,and $\gamma = -3$.
Therefore,$\alpha^2 + \beta^2 + \gamma^2 = (4)^2 + (1)^2 + (-3)^2 = 16 + 1 + 9 = 26$.

(C) Let $\vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}$ be a unit vector,so $a_1^2 + a_2^2 + a_3^2 = 1$.
Let $\vec{b} = 2 \hat{i} - \hat{j} + 2 \hat{k}$,$\vec{c} = \hat{i} + 2 \hat{j} - 2 \hat{k}$,and $\vec{d} = \hat{k}$.
The projections are equal,so $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \frac{\vec{a} \cdot \vec{c}}{|\vec{c}|} = \frac{\vec{a} \cdot \vec{d}}{|\vec{d}|}$.
Calculating magnitudes: $|\vec{b}| = \sqrt{2^2 + (-1)^2 + 2^2} = 3$,$|\vec{c}| = \sqrt{1^2 + 2^2 + (-2)^2} = 3$,and $|\vec{d}| = 1$.
Thus,$\frac{2a_1 - a_2 + 2a_3}{3} = \frac{a_1 + 2a_2 - 2a_3}{3} = a_3$.
From $\frac{2a_1 - a_2 + 2a_3}{3} = a_3$,we get $2a_1 - a_2 - a_3 = 0$.
From $\frac{a_1 + 2a_2 - 2a_3}{3} = a_3$,we get $a_1 + 2a_2 - 5a_3 = 0$.
Solving these equations,we find $a_1 = \frac{7}{5}a_3$ and $a_2 = \frac{9}{5}a_3$.
Substituting into $a_1^2 + a_2^2 + a_3^2 = 1$: $(\frac{7}{5}a_3)^2 + (\frac{9}{5}a_3)^2 + a_3^2 = 1 \implies \frac{49+81+25}{25}a_3^2 = 1 \implies a_3^2 = \frac{25}{155} \implies a_3 = \frac{5}{\sqrt{155}}$.
Thus,$\vec{a} = \frac{1}{\sqrt{155}}(7 \hat{i} + 9 \hat{j} + 5 \hat{k})$.

(C) Given $\vec{a} \times \hat{d} = \vec{b} \times \hat{d}$,we have $(\vec{a} - \vec{b}) \times \hat{d} = 0$.
This implies $\hat{d}$ is parallel to $\vec{a} - \vec{b}$.
$\vec{a} - \vec{b} = (1-3)\hat{i} + (1-2)\hat{j} + (1-(-1))\hat{k} = -2\hat{i} - \hat{j} + 2\hat{k}$.
Since $\hat{d}$ is a unit vector,$\hat{d} = \pm \frac{-2\hat{i} - \hat{j} + 2\hat{k}}{3}$.
Given $\vec{c} \cdot \vec{a} = 0$,we have $(\lambda \hat{j} + \mu \hat{k}) \cdot (\hat{i} + \hat{j} + \hat{k}) = 0$,so $\lambda + \mu = 0$,which means $\mu = -\lambda$.
Thus,$\vec{c} = \lambda(\hat{j} - \hat{k})$.
Given $\vec{c} \cdot \hat{d} = 1$,we have $\lambda(\hat{j} - \hat{k}) \cdot \pm \frac{1}{3}(-2\hat{i} - \hat{j} + 2\hat{k}) = 1$.
$\pm \frac{\lambda}{3}(-1 - 2) = 1 \Rightarrow \mp \lambda = 1 \Rightarrow \lambda = \mp 1$.
In both cases,$\lambda^2 = 1$ and $\mu^2 = \lambda^2 = 1$.
We need to calculate $|3 \lambda \hat{d} + \mu \vec{c}|^2 = 9 \lambda^2 |\hat{d}|^2 + \mu^2 |\vec{c}|^2 + 6 \lambda \mu (\hat{d} \cdot \vec{c})$.
Since $|\hat{d}|^2 = 1$,$|\vec{c}|^2 = \lambda^2 + \mu^2 = 2\lambda^2 = 2$,and $\hat{d} \cdot \vec{c} = 1$:
$|3 \lambda \hat{d} + \mu \vec{c}|^2 = 9(1)(1) + (1)(2) + 6(\lambda)(-\lambda)(1) = 9 + 2 - 6\lambda^2 = 11 - 6(1) = 5$.

(B) Given $\overrightarrow{u} = 3\hat{i} - \hat{j}$ and $\overrightarrow{v} = 2\hat{i} + \hat{j} - \lambda\hat{k}$.
The angle $\theta$ between $\overrightarrow{u}$ and $\overrightarrow{v}$ is $\cos \theta = \frac{\overrightarrow{u} \cdot \overrightarrow{v}}{|\overrightarrow{u}| |\overrightarrow{v}|}$.
$\overrightarrow{u} \cdot \overrightarrow{v} = (3)(2) + (-1)(1) + (0)(-\lambda) = 6 - 1 = 5$.
$|\overrightarrow{u}| = \sqrt{3^2 + (-1)^2} = \sqrt{10}$.
$|\overrightarrow{v}| = \sqrt{2^2 + 1^2 + (-\lambda)^2} = \sqrt{5 + \lambda^2}$.
Given $\cos \theta = \frac{\sqrt{5}}{2\sqrt{7}}$,so $\frac{5}{\sqrt{10} \sqrt{5 + \lambda^2}} = \frac{\sqrt{5}}{2\sqrt{7}}$.
Squaring both sides: $\frac{25}{10(5 + \lambda^2)} = \frac{5}{4 \times 7} = \frac{5}{28}$.
$\frac{5}{2(5 + \lambda^2)} = \frac{5}{28} \Rightarrow 5 + \lambda^2 = 14 \Rightarrow \lambda^2 = 9 \Rightarrow \lambda = 3$ (since $\lambda > 0$).
Now,$\vec{v} = \vec{v}_1 + \vec{v}_2$,where $\vec{v}_1 \parallel \overrightarrow{u}$ and $\vec{v}_2 \perp \overrightarrow{u}$.
Since $\vec{v}_1$ and $\vec{v}_2$ are orthogonal,$|\vec{v}|^2 = |\vec{v}_1|^2 + |\vec{v}_2|^2$.
$|\vec{v}|^2 = 2^2 + 1^2 + (-3)^2 = 4 + 1 + 9 = 14$.
Thus,$|\vec{v}_1|^2 + |\vec{v}_2|^2 = 14$.

(C) Given $\vec{c} = 3\hat{a} + 6\hat{b} + 9(\hat{a} \times \hat{b})$.
Since $\sin \theta = \frac{\sqrt{65}}{9}$,we have $\cos^2 \theta = 1 - \sin^2 \theta = 1 - \frac{65}{81} = \frac{16}{81}$,so $\cos \theta = \frac{4}{9}$ (as $0 < \theta < \frac{\pi}{2}$).
Now,calculate $\vec{c} \cdot \hat{a} = (3\hat{a} + 6\hat{b} + 9(\hat{a} \times \hat{b})) \cdot \hat{a} = 3(\hat{a} \cdot \hat{a}) + 6(\hat{b} \cdot \hat{a}) + 9((\hat{a} \times \hat{b}) \cdot \hat{a})$.
Since $\hat{a} \cdot \hat{a} = 1$,$\hat{b} \cdot \hat{a} = \cos \theta = \frac{4}{9}$,and $(\hat{a} \times \hat{b}) \cdot \hat{a} = 0$,we get $\vec{c} \cdot \hat{a} = 3(1) + 6(\frac{4}{9}) + 0 = 3 + \frac{8}{3} = \frac{17}{3}$.
Next,calculate $\vec{c} \cdot \hat{b} = (3\hat{a} + 6\hat{b} + 9(\hat{a} \times \hat{b})) \cdot \hat{b} = 3(\hat{a} \cdot \hat{b}) + 6(\hat{b} \cdot \hat{b}) + 9((\hat{a} \times \hat{b}) \cdot \hat{b})$.
Since $\hat{a} \cdot \hat{b} = \frac{4}{9}$,$\hat{b} \cdot \hat{b} = 1$,and $(\hat{a} \times \hat{b}) \cdot \hat{b} = 0$,we get $\vec{c} \cdot \hat{b} = 3(\frac{4}{9}) + 6(1) + 0 = \frac{4}{3} + 6 = \frac{22}{3}$.
Finally,$9(\vec{c} \cdot \hat{a}) - 3(\vec{c} \cdot \hat{b}) = 9(\frac{17}{3}) - 3(\frac{22}{3}) = 3(17) - 22 = 51 - 22 = 29$.

(C) Given $\frac{|\vec{a}+\vec{b}|+|\vec{a}-\vec{b}|}{|\vec{a}+\vec{b}|-|\vec{a}-\vec{b}|}=\sqrt{2}+1$.
Applying componendo and dividendo,we get:
$\frac{|\vec{a}+\vec{b}|}{|\vec{a}-\vec{b}|} = \frac{(\sqrt{2}+1)+1}{(\sqrt{2}+1)-1} = \frac{\sqrt{2}+2}{\sqrt{2}} = 1+\sqrt{2}$.
Squaring both sides:
$|\vec{a}+\vec{b}|^2 = (1+\sqrt{2})^2 |\vec{a}-\vec{b}|^2$.
$|\vec{a}+\vec{b}|^2 = (1+2+2\sqrt{2}) |\vec{a}-\vec{b}|^2 = (3+2\sqrt{2}) |\vec{a}-\vec{b}|^2$.
Since $|\vec{a}| = |\vec{b}|$,let $|\vec{a}|^2 = |\vec{b}|^2 = k^2$.
$2k^2 + 2\vec{a}\cdot\vec{b} = (3+2\sqrt{2})(2k^2 - 2\vec{a}\cdot\vec{b})$.
Dividing by $2k^2$:
$1 + \frac{\vec{a}\cdot\vec{b}}{k^2} = (3+2\sqrt{2})(1 - \frac{\vec{a}\cdot\vec{b}}{k^2})$.
Let $x = \frac{\vec{a}\cdot\vec{b}}{k^2}$.
$1+x = 3+2\sqrt{2} - (3+2\sqrt{2})x$.
$x(1+3+2\sqrt{2}) = 2+2\sqrt{2}$.
$x(4+2\sqrt{2}) = 2+2\sqrt{2}$.
$x = \frac{2+2\sqrt{2}}{4+2\sqrt{2}} = \frac{1+\sqrt{2}}{2+\sqrt{2}} = \frac{1}{\sqrt{2}}$.
Now,$\frac{|\vec{a}+\vec{b}|^2}{|\vec{a}|^2} = \frac{|\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a}\cdot\vec{b}}{|\vec{a}|^2} = 1 + 1 + 2x = 2 + 2(\frac{1}{\sqrt{2}}) = 2+\sqrt{2}$.

(D) Let the vector $\vec{p}$ lie in the plane of $\vec{a}$ and $\vec{b}$. Then $\vec{p} = \vec{a} + \lambda \vec{b}$.
Since $\vec{p}$ is perpendicular to $\vec{a}$,we have $\vec{p} \cdot \vec{a} = 0$.
$(\vec{a} + \lambda \vec{b}) \cdot \vec{a} = 0 \Rightarrow \vec{a} \cdot \vec{a} + \lambda (\vec{b} \cdot \vec{a}) = 0$.
Given $\vec{a} = \hat{i} + 2\hat{j} + \hat{k}$ and $\vec{b} = 2\hat{i} + \hat{j} - \hat{k}$.
$\vec{a} \cdot \vec{a} = 1^2 + 2^2 + 1^2 = 6$.
$\vec{b} \cdot \vec{a} = (2)(1) + (1)(2) + (-1)(1) = 2 + 2 - 1 = 3$.
Substituting these values: $6 + \lambda(3) = 0 \Rightarrow \lambda = -2$.
Thus,$\vec{p} = (\hat{i} + 2\hat{j} + \hat{k}) - 2(2\hat{i} + \hat{j} - \hat{k}) = -3\hat{i} + 3\hat{k}$.
The unit vector $\hat{c}$ is $\pm \frac{\vec{p}}{|\vec{p}|} = \pm \frac{-3\hat{i} + 3\hat{k}}{\sqrt{(-3)^2 + 3^2}} = \pm \frac{-3\hat{i} + 3\hat{k}}{3\sqrt{2}} = \pm \frac{-\hat{i} + \hat{k}}{\sqrt{2}}$.
Comparing with the options,the correct vector is $\frac{1}{\sqrt{2}}(-\hat{i} + \hat{k})$.

(C) In any triangle,the centroid $G$ divides the line segment joining the orthocenter $H$ and the circumcentre $P$ in the ratio $2:1$.
Thus,by the section formula,the position vector of the centroid is given by $\bar{g} = \frac{1 \cdot \bar{h} + 2 \cdot \bar{p}}{1 + 2}$.
This simplifies to $3 \bar{g} = \bar{h} + 2 \bar{p}$,which can be rewritten as $2 \bar{p} + 1 \bar{h} - 3 \bar{g} = \overline{0}$.
Comparing this with the given equation $x \bar{p} + y \bar{h} + z \bar{g} = \overline{0}$,we get $x = 2$,$y = 1$,and $z = -3$.

(C) Since $\triangle ABC$ is right-angled at $A$,the vectors $\vec{AB}$ and $\vec{AC}$ are perpendicular,so their dot product is zero: $\vec{AB} \cdot \vec{AC} = 0$.
First,we find the vectors:
$\vec{AB} = (3-4)\hat{i} + (1-2)\hat{j} + (8-x)\hat{k} = -\hat{i} - \hat{j} + (8-x)\hat{k}$.
$\vec{AC} = (2-4)\hat{i} + (-1-2)\hat{j} + (2-x)\hat{k} = -2\hat{i} - 3\hat{j} + (2-x)\hat{k}$.
Now,compute the dot product:
$(-1)(-2) + (-1)(-3) + (8-x)(2-x) = 0$.
$2 + 3 + (16 - 8x - 2x + x^2) = 0$.
$5 + 16 - 10x + x^2 = 0$.
$x^2 - 10x + 21 = 0$.
Factoring the quadratic equation:
$(x-3)(x-7) = 0$.
Thus,$x = 3$ or $x = 7$.
Given the options,the correct value is $3$.

(B) Given: $|\overline{a}|=2, |\overline{b}|=4, |\overline{c}|=4$.
According to the condition,the projection of $\overline{b}$ on $\overline{a}$ is equal to the projection of $\overline{c}$ on $\overline{a}$.
$\frac{\overline{b} \cdot \overline{a}}{|\overline{a}|} = \frac{\overline{c} \cdot \overline{a}}{|\overline{a}|}$
$\implies \overline{b} \cdot \overline{a} = \overline{c} \cdot \overline{a}$
$\implies (\overline{b} - \overline{c}) \cdot \overline{a} = 0$ ... $(i)$
Also,$\overline{b}$ is perpendicular to $\overline{c}$,so $\overline{b} \cdot \overline{c} = 0$.
Now,consider $|\overline{a} + \overline{b} - \overline{c}|^2 = |\overline{a}|^2 + |\overline{b} - \overline{c}|^2 + 2\overline{a} \cdot (\overline{b} - \overline{c})$.
From $(i)$,$\overline{a} \cdot (\overline{b} - \overline{c}) = 0$.
So,$|\overline{a} + \overline{b} - \overline{c}|^2 = |\overline{a}|^2 + |\overline{b}|^2 + |\overline{c}|^2 - 2(\overline{b} \cdot \overline{c})$.
Since $\overline{b} \cdot \overline{c} = 0$,we have:
$|\overline{a} + \overline{b} - \overline{c}|^2 = (2)^2 + (4)^2 + (4)^2 - 0 = 4 + 16 + 16 = 36$.
Therefore,$|\overline{a} + \overline{b} - \overline{c}| = \sqrt{36} = 6$.

(B) First,we find the vectors $\overline{AB}$ and $\overline{CD}$:
$\overline{AB} = (1-2)\hat{i} + (-4 - (-3))\hat{j} + (-2-0)\hat{k} = -\hat{i} - \hat{j} - 2\hat{k}$
$\overline{CD} = (7-4)\hat{i} + (0-6)\hat{j} + (10-8)\hat{k} = 3\hat{i} - 6\hat{j} + 2\hat{k}$
The scalar projection of vector $\overline{AB}$ on vector $\overline{CD}$ is given by the formula:
$\text{Projection} = \frac{\overline{AB} \cdot \overline{CD}}{|\overline{CD}|}$
Calculate the dot product $\overline{AB} \cdot \overline{CD}$:
$\overline{AB} \cdot \overline{CD} = (-1)(3) + (-1)(-6) + (-2)(2) = -3 + 6 - 4 = -1$
Calculate the magnitude of $\overline{CD}$:
$|\overline{CD}| = \sqrt{3^2 + (-6)^2 + 2^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7$
Thus,the scalar projection is:
$\frac{-1}{7}$
Note: The magnitude of the projection is $|\frac{-1}{7}| = \frac{1}{7}$.

(C) The vector projection of $\vec{b}$ on $\vec{a}$ is given by the formula: $\left(\frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2}\right) \vec{a}$.
First,calculate the dot product $\vec{b} \cdot \vec{a} = (7)(3) + (-5)(2) + (-1)(5) = 21 - 10 - 5 = 6$.
Next,calculate the square of the magnitude of $\vec{a}$: $|\vec{a}|^2 = 3^2 + 2^2 + 5^2 = 9 + 4 + 25 = 38$.
Now,substitute these values into the formula:
Vector projection = $\frac{6}{38} (3 \hat{i} + 2 \hat{j} + 5 \hat{k}) = \frac{3}{19} (3 \hat{i} + 2 \hat{j} + 5 \hat{k})$.

(B) Given points are $A=(-2,2,3), B=(3,2,2), C=(4,-3,5)$,and $D=(7,-5,-1)$.
The vector $\overline{AB} = (3 - (-2))\hat{i} + (2 - 2)\hat{j} + (2 - 3)\hat{k} = 5\hat{i} + 0\hat{j} - 1\hat{k} = 5\hat{i} - \hat{k}$.
The vector $\overline{CD} = (7 - 4)\hat{i} + (-5 - (-3))\hat{j} + (-1 - 5)\hat{k} = 3\hat{i} - 2\hat{j} - 6\hat{k}$.
The projection of vector $\overline{AB}$ on vector $\overline{CD}$ is given by the formula $\frac{\overline{AB} \cdot \overline{CD}}{|\overline{CD}|}$.
First,calculate the dot product: $\overline{AB} \cdot \overline{CD} = (5)(3) + (0)(-2) + (-1)(-6) = 15 + 0 + 6 = 21$.
Next,calculate the magnitude of $\overline{CD}$: $|\overline{CD}| = \sqrt{(3)^2 + (-2)^2 + (-6)^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$.
Therefore,the projection is $\frac{21}{7} = 3$.

(A) Any vector $\overrightarrow{r}$ in the plane of $\overrightarrow{b}$ and $\overrightarrow{c}$ can be written as $\overrightarrow{r} = m\overrightarrow{b} + n\overrightarrow{c}$. For simplicity,consider $\overrightarrow{r} = m\overrightarrow{b} + \overrightarrow{c}$ (assuming the vector is not parallel to $\overrightarrow{c}$).
$\overrightarrow{r} = m(\hat{i} + 2\hat{j} - \hat{k}) + (\hat{i} + \hat{j} - 2\hat{k}) = (m+1)\hat{i} + (2m+1)\hat{j} + (-m-2)\hat{k}$.
The projection of $\overrightarrow{r}$ on $\overrightarrow{a}$ is given by $\frac{|\overrightarrow{r} \cdot \overrightarrow{a}|}{|\overrightarrow{a}|} = \sqrt{\frac{2}{3}}$.
$\overrightarrow{a} = 2\hat{i} - \hat{j} + \hat{k}$,so $|\overrightarrow{a}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{6}$.
$\overrightarrow{r} \cdot \overrightarrow{a} = 2(m+1) - 1(2m+1) + 1(-m-2) = 2m + 2 - 2m - 1 - m - 2 = -m - 1$.
Thus,$\frac{|-m-1|}{\sqrt{6}} = \sqrt{\frac{2}{3}} = \frac{\sqrt{2}}{\sqrt{3}} = \frac{2}{\sqrt{6}}$.
$|-m-1| = 2$,which implies $-m-1 = 2$ or $-m-1 = -2$.
If $-m-1 = 2$,then $m = -3$. Substituting $m = -3$ into $\overrightarrow{r}$: $\overrightarrow{r} = -2\hat{i} - 5\hat{j} + \hat{k}$.
If $-m-1 = -2$,then $m = 1$. Substituting $m = 1$ into $\overrightarrow{r}$: $\overrightarrow{r} = 2\hat{i} + 3\hat{j} - 3\hat{k}$.
Comparing with the options,$2\hat{i} + 3\hat{j} - 3\hat{k}$ is present.

(B) Given the equations of the lines are $\bar{r} \times \bar{a} = \bar{b} \times \bar{a}$ and $\bar{r} \times \bar{b} = \bar{a} \times \bar{b}$.
These can be rewritten as $\bar{r} \times \bar{a} - \bar{b} \times \bar{a} = 0$,which implies $(\bar{r} - \bar{b}) \times \bar{a} = 0$. This means $\bar{r} - \bar{b} = t\bar{a}$ for some scalar $t$,so $\bar{r} = \bar{b} + t\bar{a}$.
Similarly,$\bar{r} \times \bar{b} - \bar{a} \times \bar{b} = 0$ implies $(\bar{r} - \bar{a}) \times \bar{b} = 0$,so $\bar{r} = \bar{a} + s\bar{b}$ for some scalar $s$.
Equating the two expressions for $\bar{r}$: $\bar{b} + t\bar{a} = \bar{a} + s\bar{b}$.
Substituting $\bar{a} = \hat{i} + \hat{j}$ and $\bar{b} = 2\hat{i} - \hat{k}$:
$(2\hat{i} - \hat{k}) + t(\hat{i} + \hat{j}) = (\hat{i} + \hat{j}) + s(2\hat{i} - \hat{k})$.
Comparing components:
$i: 2 + t = 1 + 2s \implies t - 2s = -1$
$j: t = 1$
$k: -1 = -s \implies s = 1$
Substituting $t=1$ and $s=1$ into the equation for $\bar{r}$:
$\bar{r} = \bar{b} + 1\bar{a} = (2\hat{i} - \hat{k}) + (\hat{i} + \hat{j}) = 3\hat{i} + \hat{j} - \hat{k}$.
Thus,the point of intersection is $(3, 1, -1)$.

(A) Given that $\triangle PQR$ is right-angled at $Q$.
Therefore,the vectors $\vec{QP}$ and $\vec{QR}$ are perpendicular to each other,which means their dot product is zero: $\vec{QP} \cdot \vec{QR} = 0$.
First,find the vectors $\vec{QP}$ and $\vec{QR}$:
$\vec{QP} = (6-1, 10-0, 10-(-5)) = (5, 10, 15)$
$\vec{QR} = (6-1, -10-0, \lambda-(-5)) = (5, -10, \lambda+5)$
Now,calculate the dot product:
$\vec{QP} \cdot \vec{QR} = (5)(5) + (10)(-10) + (15)(\lambda+5) = 0$
$25 - 100 + 15\lambda + 75 = 0$
$-75 + 15\lambda + 75 = 0$
$15\lambda = 0$
$\lambda = 0$

(C) Let the position vectors of vertices $A, B, C$ be $\vec{a}, \vec{b}, \vec{c}$ respectively.
Since $P$ divides $BC$ in ratio $3:4$,the position vector of $P$ is $\vec{p} = \frac{4\vec{b} + 3\vec{c}}{7}$.
Since $Q$ divides $CA$ in ratio $5:3$,the position vector of $Q$ is $\vec{q} = \frac{3\vec{c} + 5\vec{a}}{8}$.
Let $G$ divide $AP$ in ratio $k:1$. Then $\vec{g} = \frac{1\vec{a} + k\vec{p}}{k+1} = \frac{\vec{a} + k(\frac{4\vec{b} + 3\vec{c}}{7})}{k+1} = \frac{7\vec{a} + 4k\vec{b} + 3k\vec{c}}{7(k+1)}$.
Also,$G$ lies on $BQ$,so $\vec{g} = \frac{m\vec{q} + 1\vec{b}}{m+1} = \frac{m(\frac{3\vec{c} + 5\vec{a}}{8}) + \vec{b}}{m+1} = \frac{5m\vec{a} + 8\vec{b} + 3m\vec{c}}{8(m+1)}$.
Comparing coefficients of $\vec{a}, \vec{b}, \vec{c}$:
$\frac{7}{7(k+1)} = \frac{5m}{8(m+1)} \implies \frac{1}{k+1} = \frac{5m}{8(m+1)}$
$\frac{4k}{7(k+1)} = \frac{8}{8(m+1)} \implies \frac{4k}{7(k+1)} = \frac{1}{m+1}$
$\frac{3k}{7(k+1)} = \frac{3m}{8(m+1)} \implies \frac{k}{7(k+1)} = \frac{m}{8(m+1)}$
From the first and third equations,$\frac{1}{k+1} = 5 \times \frac{k}{7(k+1)} \implies 1 = \frac{5k}{7} \implies k = \frac{7}{5}$.
Thus,the ratio is $7:5$.

(B) Given that $\bar{a}$ and $\bar{b}$ are unit vectors,we have $|\bar{a}| = 1$ and $|\bar{b}| = 1$.
Since $\bar{a}+\bar{b}$ is a unit vector,$|\bar{a}+\bar{b}| = 1$.
Squaring both sides,we get $|\bar{a}+\bar{b}|^2 = 1^2$.
Using the property $|\bar{x}|^2 = \bar{x} \cdot \bar{x}$,we have $(\bar{a}+\bar{b}) \cdot (\bar{a}+\bar{b}) = 1$.
Expanding the dot product,we get $|\bar{a}|^2 + |\bar{b}|^2 + 2(\bar{a} \cdot \bar{b}) = 1$.
Substituting the values $|\bar{a}| = 1$ and $|\bar{b}| = 1$,we get $1^2 + 1^2 + 2|\bar{a}||\bar{b}| \cos \theta = 1$.
$1 + 1 + 2(1)(1) \cos \theta = 1$.
$2 + 2 \cos \theta = 1$.
$2 \cos \theta = -1$.
$\cos \theta = -\frac{1}{2}$.
Since $\cos \theta = -\frac{1}{2}$,the angle $\theta = \frac{2 \pi}{3}$.

(D) Since $\bar{c}$ is coplanar with $\bar{a}$ and $\bar{b}$,we can write $\bar{c} = x\bar{a} + y\bar{b}$ for some scalars $x$ and $y$.
$\bar{c} = x(2\hat{i} + \hat{j} + \hat{k}) + y(\hat{i} + 2\hat{j} - \hat{k}) = (2x + y)\hat{i} + (x + 2y)\hat{j} + (x - y)\hat{k}$.
Given that $\bar{c} \perp \bar{a}$,the dot product $\bar{c} \cdot \bar{a} = 0$.
$(2x + y)(2) + (x + 2y)(1) + (x - y)(1) = 0$.
$4x + 2y + x + 2y + x - y = 0 \implies 6x + 3y = 0 \implies y = -2x$.
Substituting $y = -2x$ into the expression for $\bar{c}$:
$\bar{c} = x(2\hat{i} + \hat{j} + \hat{k}) - 2x(\hat{i} + 2\hat{j} - \hat{k}) = x(2\hat{i} + \hat{j} + \hat{k} - 2\hat{i} - 4\hat{j} + 2\hat{k}) = x(-3\hat{j} + 3\hat{k}) = 3x(-\hat{j} + \hat{k})$.
For $x = 1/3$,we get $\bar{c} = -\hat{j} + \hat{k}$. Thus,option $D$ is correct.

(C) Let the position vectors of vertices $A, B, C$ be $\vec{a}, \vec{b}, \vec{c}$ respectively.
$P$ divides $AC$ in ratio $3:4$,so $\vec{p} = \frac{4\vec{a} + 3\vec{c}}{7}$.
$Q$ divides $BC$ in ratio $4:3$,so $\vec{q} = \frac{3\vec{b} + 4\vec{c}}{7}$.
$M$ is the intersection of $BP$ and $AQ$. Let $M$ divide $AQ$ in ratio $k:1$. Then $\vec{m} = \frac{k\vec{q} + \vec{a}}{k+1} = \frac{k(\frac{3\vec{b} + 4\vec{c}}{7}) + \vec{a}}{k+1} = \frac{7\vec{a} + 3k\vec{b} + 4k\vec{c}}{7(k+1)}$.
Also,$M$ lies on $BP$,so $\vec{m} = (1-t)\vec{b} + t\vec{p} = (1-t)\vec{b} + t(\frac{4\vec{a} + 3\vec{c}}{7}) = \frac{4t\vec{a} + 7(1-t)\vec{b} + 3t\vec{c}}{7}$.
Comparing coefficients of $\vec{a}, \vec{b}, \vec{c}$:
$\frac{7}{7(k+1)} = \frac{4t}{7} \implies 1 = \frac{4t(k+1)}{7} \implies 4t(k+1) = 7$.
$\frac{3k}{7(k+1)} = 1-t \implies 3k = 7(1-t)(k+1) = 7(k+1-tk-t) = 7k+7-7tk-7t$.
Substitute $4tk = 7-4t$ from the first equation:
$3k = 7k+7 - (7-4t) - 7t = 7k+7-7+4t-7t = 7k-3t$.
$3t = 4k \implies t = \frac{4k}{3}$.
Substitute $t$ into $4t(k+1) = 7$:
$4(\frac{4k}{3})(k+1) = 7 \implies 16k^2 + 16k = 21 \implies 16k^2 + 16k - 21 = 0$.
$(4k+7)(4k-3) = 0$. Since $k>0$,$k = \frac{3}{4}$.
Thus,the ratio is $3:4$. Wait,checking the options,let's re-evaluate. The ratio $M$ divides $AQ$ is $k:1 = 3/4 : 1 = 3:4$. None of the options match. Re-reading: $P$ on $AC$ $(3:4)$,$Q$ on $BC$ $(4:3)$. Using Menelaus theorem on $\triangle ACQ$ with line $B-M-P$: $\frac{AM}{MQ} \cdot \frac{QB}{BC} \cdot \frac{CP}{PA} = 1$. $\frac{AM}{MQ} \cdot \frac{3}{7} \cdot \frac{4}{3} = 1 \implies \frac{AM}{MQ} = \frac{7}{4}$. The ratio is $7:4$. Still not matching. Let's re-calculate: $P$ divides $AC$ as $AP:PC = 3:4$. $Q$ divides $BC$ as $BQ:QC = 4:3$. By Menelaus on $\triangle ACQ$ with line $B-M-P$: $\frac{AM}{MQ} \cdot \frac{QB}{BC} \cdot \frac{CP}{PA} = 1$. $\frac{AM}{MQ} \cdot \frac{4}{7} \cdot \frac{4}{3} = 1 \implies \frac{AM}{MQ} = \frac{21}{16}$. This matches option $C$.

(C) Let $\vec{a} = 4\hat{i}+7\hat{j}+8\hat{k}$,$\vec{b} = 2\hat{i}+3\hat{j}+4\hat{k}$,and $\vec{c} = 2\hat{i}+5\hat{j}+7\hat{k}$.
By the Angle Bisector Theorem,the bisector of $\angle B$ divides the opposite side $AC$ in the ratio $BA : BC$.
Calculate the lengths $BA$ and $BC$:
$BA = |\vec{a} - \vec{b}| = |(4-2)\hat{i} + (7-3)\hat{j} + (8-4)\hat{k}| = |2\hat{i} + 4\hat{j} + 4\hat{k}| = \sqrt{2^2 + 4^2 + 4^2} = \sqrt{4+16+16} = \sqrt{36} = 6$.
$BC = |\vec{c} - \vec{b}| = |(2-2)\hat{i} + (5-3)\hat{j} + (7-4)\hat{k}| = |0\hat{i} + 2\hat{j} + 3\hat{k}| = \sqrt{0^2 + 2^2 + 3^2} = \sqrt{4+9} = \sqrt{13}$.
The point $D$ on $AC$ divides $AC$ in the ratio $BA : BC = 6 : \sqrt{13}$.
Using the section formula,the position vector $\vec{d}$ is given by $\vec{d} = \frac{6\vec{c} + \sqrt{13}\vec{a}}{6+\sqrt{13}}$.
$\vec{d} = \frac{6(2\hat{i}+5\hat{j}+7\hat{k}) + \sqrt{13}(4\hat{i}+7\hat{j}+8\hat{k})}{6+\sqrt{13}} = \frac{(12+4\sqrt{13})\hat{i} + (30+7\sqrt{13})\hat{j} + (42+8\sqrt{13})\hat{k}}{6+\sqrt{13}}$.

(B) The area of a quadrilateral $ABCD$ can be calculated as the sum of the areas of two triangles,$\triangle ABC$ and $\triangle ADC$.
Area of $\triangle ABC = \frac{1}{2} |\overline{AB} \times \overline{AC}| = \frac{1}{2} |\bar{a} \times (2\bar{a} + 3\bar{b})| = \frac{1}{2} |2(\bar{a} \times \bar{a}) + 3(\bar{a} \times \bar{b})| = \frac{1}{2} |0 + 3(\bar{a} \times \bar{b})| = \frac{3}{2} |\bar{a} \times \bar{b}|$.
Area of $\triangle ADC = \frac{1}{2} |\overline{AD} \times \overline{AC}| = \frac{1}{2} |\bar{b} \times (2\bar{a} + 3\bar{b})| = \frac{1}{2} |2(\bar{b} \times \bar{a}) + 3(\bar{b} \times \bar{b})| = \frac{1}{2} |-2(\bar{a} \times \bar{b}) + 0| = |\bar{a} \times \bar{b}|$.
Total area of quadrilateral $ABCD = \frac{3}{2} |\bar{a} \times \bar{b}| + |\bar{a} \times \bar{b}| = \frac{5}{2} |\bar{a} \times \bar{b}|$.
The area of the parallelogram with adjacent sides $AB$ and $AD$ is $|\bar{a} \times \bar{b}|$.
Given that the area of the quadrilateral is $\alpha$ times the area of the parallelogram,we have $\frac{5}{2} |\bar{a} \times \bar{b}| = \alpha |\bar{a} \times \bar{b}|$.
Thus,$\alpha = \frac{5}{2}$.

(A) Let $\vec{r}$ and $\vec{s}$ be the position vectors of points $R$ and $S$ respectively.
Using the section formula for internal division in ratio $2:3$:
$\vec{r} = \frac{2\vec{q} + 3\vec{p}}{2+3} = \frac{3\vec{p} + 2\vec{q}}{5}$
Using the section formula for external division in ratio $2:3$:
$\vec{s} = \frac{2\vec{q} - 3\vec{p}}{2-3} = \frac{2\vec{q} - 3\vec{p}}{-1} = 3\vec{p} - 2\vec{q}$
Since $\vec{OR}$ and $\vec{OS}$ are perpendicular,their dot product is zero:
$\vec{r} \cdot \vec{s} = 0$
$\left(\frac{3\vec{p} + 2\vec{q}}{5}\right) \cdot (3\vec{p} - 2\vec{q}) = 0$
$(3\vec{p} + 2\vec{q}) \cdot (3\vec{p} - 2\vec{q}) = 0$
$9|\vec{p}|^2 - 6(\vec{p} \cdot \vec{q}) + 6(\vec{q} \cdot \vec{p}) - 4|\vec{q}|^2 = 0$
$9p^2 - 4q^2 = 0$
$9p^2 = 4q^2$

(A) Since $\overline{a}$ bisects the angle between $\overline{b}$ and $\overline{c}$,it must be proportional to the sum of the unit vectors along $\overline{b}$ and $\overline{c}$.
Let $\hat{b} = \frac{\hat{i}+\hat{j}}{\sqrt{2}}$ and $\hat{c} = \frac{\hat{j}+\hat{k}}{\sqrt{2}}$.
The vector $\overline{a}$ is given by $\overline{a} = k(\hat{b} + \hat{c})$ for some scalar $k$.
$\overline{a} = k \left( \frac{\hat{i}+\hat{j}}{\sqrt{2}} + \frac{\hat{j}+\hat{k}}{\sqrt{2}} \right) = \frac{k}{\sqrt{2}} (\hat{i} + 2\hat{j} + \hat{k})$.
Given $\overline{a} = \alpha \hat{i} + 2 \hat{j} + \beta \hat{k}$.
Comparing the components,we have $\alpha = \frac{k}{\sqrt{2}}$,$2 = \frac{2k}{\sqrt{2}}$,and $\beta = \frac{k}{\sqrt{2}}$.
From $2 = \frac{2k}{\sqrt{2}}$,we get $k = \sqrt{2}$.
Substituting $k = \sqrt{2}$ into the expressions for $\alpha$ and $\beta$,we get $\alpha = \frac{\sqrt{2}}{\sqrt{2}} = 1$ and $\beta = \frac{\sqrt{2}}{\sqrt{2}} = 1$.
Thus,$\alpha=1$ and $\beta=1$.

(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,$\hat{u} = \frac{\frac{1}{\sqrt{2}}(\hat{a} + \hat{b})}{|\frac{1}{\sqrt{2}}(\hat{a} + \hat{b})|} = \frac{\hat{a} + \hat{b}}{|\hat{a} + \hat{b}|}$.

(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,$\overline{a}+2\overline{b}+6\overline{c} = -6\overline{c} + 6\overline{c} = \overline{0}$.

(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) Given $\overline{OP} = \hat{a} \sin t + \hat{b} \cos t$.
$M = |\overline{OP}| = \sqrt{(\hat{a} \sin t + \hat{b} \cos t)^2}$.
Since $\hat{a}$ and $\hat{b}$ are unit vectors,$|\hat{a}| = 1$ and $|\hat{b}| = 1$.
$M^2 = \sin^2 t |\hat{a}|^2 + \cos^2 t |\hat{b}|^2 + 2 \sin t \cos t (\hat{a} \cdot \hat{b}) = \sin^2 t + \cos^2 t + \sin 2t (\hat{a} \cdot \hat{b}) = 1 + \sin 2t (\hat{a} \cdot \hat{b})$.
For $P$ to be farthest from $O$,$M$ must be maximum.
Since $\hat{a}$ and $\hat{b}$ form an acute angle,$\hat{a} \cdot \hat{b} > 0$.
Thus,$M$ is maximum when $\sin 2t = 1$,which implies $2t = \frac{\pi}{2}$,so $t = \frac{\pi}{4}$.
At $t = \frac{\pi}{4}$,$\sin t = \cos t = \frac{1}{\sqrt{2}}$.
$M = \sqrt{1 + 1(\hat{a} \cdot \hat{b})} = (1 + \hat{a} \cdot \hat{b})^{\frac{1}{2}}$.
$\hat{u} = \frac{\overline{OP}}{|\overline{OP}|} = \frac{\hat{a} \sin t + \hat{b} \cos t}{M} = \frac{\hat{a} \frac{1}{\sqrt{2}} + \hat{b} \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}} |\hat{a} + \hat{b}|} = \frac{\hat{a} + \hat{b}}{|\hat{a} + \hat{b}|}$.

(D) According to the given condition,the vector $(\bar{a}+\lambda \bar{b})$ is perpendicular to $\bar{c}$,so their dot product must be zero:
$(\bar{a}+\lambda \bar{b}) \cdot \bar{c} = 0$
First,calculate $\bar{a}+\lambda \bar{b}$:
$\bar{a}+\lambda \bar{b} = (2 \hat{i}+3 \hat{j}+2 \hat{k}) + \lambda(2 \hat{i}+\hat{j}-\hat{k}) = (2+2\lambda) \hat{i} + (3+\lambda) \hat{j} + (2-\lambda) \hat{k}$
Now,take the dot product with $\bar{c} = 3 \hat{i} - \hat{j} + 0 \hat{k}$:
$[(2+2\lambda) \hat{i} + (3+\lambda) \hat{j} + (2-\lambda) \hat{k}] \cdot (3 \hat{i} - \hat{j} + 0 \hat{k}) = 0$
$3(2+2\lambda) - 1(3+\lambda) + 0(2-\lambda) = 0$
$6 + 6\lambda - 3 - \lambda = 0$
$3 + 5\lambda = 0$
$5\lambda = -3$
$\lambda = \frac{-3}{5}$

(C) Given: $|\overline{u}|=1, |\overline{v}|=2, |\overline{w}|=3$.
According to the condition,the projection of $\overline{v}$ along $\overline{u}$ is equal to the projection of $\overline{w}$ along $\overline{u}$.
$\frac{\overline{v} \cdot \overline{u}}{|\overline{u}|} = \frac{\overline{w} \cdot \overline{u}}{|\overline{u}|}$
$\implies \overline{v} \cdot \overline{u} = \overline{w} \cdot \overline{u}$
$\implies (\overline{w} - \overline{v}) \cdot \overline{u} = 0$.
Now,consider $|\overline{u} - \overline{v} + \overline{w}|^2 = |\overline{u} + (\overline{w} - \overline{v})|^2$.
$= |\overline{u}|^2 + |\overline{w} - \overline{v}|^2 + 2\overline{u} \cdot (\overline{w} - \overline{v})$.
Since $(\overline{w} - \overline{v}) \cdot \overline{u} = 0$,the last term is $0$.
$= |\overline{u}|^2 + |\overline{w}|^2 + |\overline{v}|^2 - 2(\overline{w} \cdot \overline{v})$.
Since $\overline{v}$ and $\overline{w}$ are perpendicular,$\overline{w} \cdot \overline{v} = 0$.
$= (1)^2 + (3)^2 + (2)^2 - 0 = 1 + 9 + 4 = 14$.
Therefore,$|\overline{u} - \overline{v} + \overline{w}| = \sqrt{14}$.

(D) Given $|\overline{b}|=4, |\overline{c}|=1$ and $|\overline{b} \times \overline{c}|=\sqrt{15}$.
Let $\alpha$ be the angle between $\overline{b}$ and $\overline{c}$.
$|\overline{b} \times \overline{c}| = |\overline{b}||\overline{c}| \sin \alpha = \sqrt{15}$.
$4 \times 1 \times \sin \alpha = \sqrt{15} \implies \sin \alpha = \frac{\sqrt{15}}{4}$.
Then,$\cos^2 \alpha = 1 - \sin^2 \alpha = 1 - \frac{15}{16} = \frac{1}{16} \implies \cos \alpha = \frac{1}{4}$ (assuming $\alpha$ is acute).
Given $\overline{b} = 2\overline{c} + \lambda \overline{a}$,we have $\overline{b} - 2\overline{c} = \lambda \overline{a}$.
Squaring both sides:
$|\overline{b} - 2\overline{c}|^2 = |\lambda \overline{a}|^2$.
$|\overline{b}|^2 + 4|\overline{c}|^2 - 4(\overline{b} \cdot \overline{c}) = \lambda^2 |\overline{a}|^2$.
$|\overline{b}|^2 + 4|\overline{c}|^2 - 4(|\overline{b}||\overline{c}| \cos \alpha) = \lambda^2 |\overline{a}|^2$.
$16 + 4(1) - 4(4 \times 1 \times \frac{1}{4}) = \lambda^2 (2)^2$.
$16 + 4 - 4 = 4\lambda^2$.
$16 = 4\lambda^2 \implies \lambda^2 = 4 \implies \lambda = \pm 2$.
Since $4$ is an option,the value is $4$.