Basic , Modulus and Algebra of vectors Questions in English

(B) Given the equation: $\left(\frac{a_1}{2} + \frac{a_2}{4} + \frac{a_3}{8}\right)^2 = \frac{a_1^2}{2} + \frac{a_2^2}{4} + \frac{a_3^2}{8}$.
Expanding the left side: $\frac{a_1^2}{4} + \frac{a_2^2}{16} + \frac{a_3^2}{64} + \frac{a_1 a_2}{4} + \frac{a_2 a_3}{16} + \frac{a_3 a_1}{8} = \frac{a_1^2}{2} + \frac{a_2^2}{4} + \frac{a_3^2}{8}$.
Multiplying by $64$ to clear denominators: $16a_1^2 + 4a_2^2 + a_3^2 + 16a_1 a_2 + 4a_2 a_3 + 8a_3 a_1 = 32a_1^2 + 16a_2^2 + 8a_3^2$.
Rearranging terms: $16a_1^2 + 12a_2^2 + 7a_3^2 - 16a_1 a_2 - 4a_2 a_3 - 8a_3 a_1 = 0$.
This can be rewritten as: $8(a_1 - a_2)^2 + (2a_2 - a_3)^2 + 2(a_3 - 2a_1)^2 + 4a_3^2 = 0$.
Since the sum of squares is zero,each term must be zero: $a_1 - a_2 = 0$,$2a_2 - a_3 = 0$,$a_3 - 2a_1 = 0$,and $a_3 = 0$.
This implies $a_1 = a_2 = a_3 = 0$.
Thus,the set $A$ contains exactly one element,which is the zero vector $(0, 0, 0)$.

(C) Given $|\vec{v} - \hat{i}| = |\vec{v} - 2\hat{j}| = |\vec{v} - \hat{j}|$.
Let $\vec{v} = x\hat{i} + y\hat{j}$. The points are $A(1, 0)$,$B(0, 2)$,and $C(0, 1)$.
Since $\vec{v}$ is equidistant from $A, B$,and $C$,it is the circumcenter of $\triangle ABC$.
Equating the squared distances:
$|\vec{v} - \hat{i}|^2 = |\vec{v} - \hat{j}|^2 \Rightarrow (x-1)^2 + y^2 = x^2 + (y-1)^2$
$x^2 - 2x + 1 + y^2 = x^2 + y^2 - 2y + 1 \Rightarrow x = y$.
Equating $|\vec{v} - \hat{j}|^2 = |\vec{v} - 2\hat{j}|^2$:
$x^2 + (y-1)^2 = x^2 + (y-2)^2$
$y^2 - 2y + 1 = y^2 - 4y + 4 \Rightarrow 2y = 3 \Rightarrow y = \frac{3}{2}$.
Since $x = y$,we have $x = \frac{3}{2}$.
Thus,$\vec{v} = \frac{3}{2}\hat{i} + \frac{3}{2}\hat{j}$.
$|\vec{v}| = \sqrt{(\frac{3}{2})^2 + (\frac{3}{2})^2} = \sqrt{\frac{9}{4} + \frac{9}{4}} = \sqrt{\frac{18}{4}} = \sqrt{4.5} \approx 2.12$.
Since $2 < 2.12 \leq 3$,$|\vec{v}|$ lies in the interval $(2, 3]$.

(C) Let the position vectors of vertices $A, B, C$ be $\vec{a}, \vec{b}, \vec{c}$ respectively. Let the origin be at the circumcenter $O$,so $\vec{O} = \vec{0}$.
Then the position vector of the orthocenter $H$ is $\vec{H} = \vec{a} + \vec{b} + \vec{c}$.
We need to evaluate $\vec{HA} + \vec{HB} + \vec{HC}$.
$\vec{HA} + \vec{HB} + \vec{HC} = (\vec{a} - \vec{H}) + (\vec{b} - \vec{H}) + (\vec{c} - \vec{H})$
$= (\vec{a} + \vec{b} + \vec{c}) - 3\vec{H}$
$= \vec{H} - 3\vec{H}$
$= -2\vec{H}$
Since $\vec{O} = \vec{0}$,$\vec{HO} = \vec{O} - \vec{H} = -\vec{H}$.
Therefore,$-2\vec{H} = 2(-\vec{H}) = 2\vec{HO}$.
Thus,$\vec{HA} + \vec{HB} + \vec{HC} = 2\vec{HO}$.

(B) Given vectors are $a = 3 \hat{i} - 4 \hat{k}$ and $b = 5 \hat{j} + 12 \hat{k}$.
The magnitudes of the vectors are:
$|a| = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$
$|b| = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$
The unit vectors along $a$ and $b$ are $\hat{a} = \frac{a}{|a|} = \frac{3 \hat{i} - 4 \hat{k}}{5}$ and $\hat{b} = \frac{b}{|b|} = \frac{5 \hat{j} + 12 \hat{k}}{13}$.
$A$ vector that bisects the angle between $a$ and $b$ is given by the sum of the unit vectors: $\hat{a} + \hat{b}$.
$\hat{a} + \hat{b} = \frac{3 \hat{i} - 4 \hat{k}}{5} + \frac{5 \hat{j} + 12 \hat{k}}{13}$
Taking the common denominator $65$:
$= \frac{13(3 \hat{i} - 4 \hat{k}) + 5(5 \hat{j} + 12 \hat{k})}{65}$
$= \frac{39 \hat{i} - 52 \hat{k} + 25 \hat{j} + 60 \hat{k}}{65}$
$= \frac{39 \hat{i} + 25 \hat{j} + 8 \hat{k}}{65}$
Since any scalar multiple of this vector also bisects the angle,we can multiply by $65$ to get the vector $39 \hat{i} + 25 \hat{j} + 8 \hat{k}$.

(A) Let the points be $A(\alpha, 10, 13)$,$B(6, 11, 11)$,and $C(\frac{9}{2}, \beta, -8)$.
Since the points are collinear,the vectors $\vec{AB}$ and $\vec{BC}$ must be parallel.
$\vec{AB} = (6 - \alpha) \hat{i} + (11 - 10) \hat{j} + (11 - 13) \hat{k} = (6 - \alpha) \hat{i} + 1 \hat{j} - 2 \hat{k}$.
$\vec{BC} = (\frac{9}{2} - 6) \hat{i} + (\beta - 11) \hat{j} + (-8 - 11) \hat{k} = -\frac{3}{2} \hat{i} + (\beta - 11) \hat{j} - 19 \hat{k}$.
Since $\vec{AB} = k \vec{BC}$,we have:
$\frac{6 - \alpha}{-3/2} = \frac{1}{\beta - 11} = \frac{-2}{-19} = \frac{2}{19}$.
From $\frac{1}{\beta - 11} = \frac{2}{19}$,we get $2(\beta - 11) = 19 \implies 2\beta - 22 = 19 \implies 2\beta = 41$.
From $\frac{6 - \alpha}{-3/2} = \frac{2}{19}$,we get $6 - \alpha = \frac{2}{19} \times (-\frac{3}{2}) = -\frac{3}{19} \implies \alpha = 6 + \frac{3}{19} = \frac{114 + 3}{19} = \frac{117}{19}$.
Now,calculate $(19 \alpha - 6 \beta)^2 = (19 \times \frac{117}{19} - 3 \times 2\beta)^2 = (117 - 3 \times 41)^2 = (117 - 123)^2 = (-6)^2 = 36$.

(D) Without loss of generality,let $|a_1| \leq |a_2| \leq |a_3|$.
For statement $(A)$:
$|\vec{a}|^2 = |a_1|^2 + |a_2|^2 + |a_3|^2 \geq |a_3|^2$.
Taking the square root,$|\vec{a}| \geq |a_3| = \max \{|a_1|, |a_2|, |a_3|\}$.
Thus,$(A)$ is true.
For statement $(B)$:
$|\vec{a}|^2 = |a_1|^2 + |a_2|^2 + |a_3|^2 \leq |a_3|^2 + |a_3|^2 + |a_3|^2 = 3|a_3|^2$.
Taking the square root,$|\vec{a}| \leq \sqrt{3} |a_3| = \sqrt{3} \max \{|a_1|, |a_2|, |a_3|\}$.
Since $\sqrt{3} < 3$,it follows that $|\vec{a}| \leq 3 \max \{|a_1|, |a_2|, |a_3|\}$.
Thus,$(B)$ is also true.
Therefore,both $(A)$ and $(B)$ are true.

(C) Let the position vectors of vertices $A, B, C, D$ be $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}, \overrightarrow{d}$ respectively.
Since $E$ is the midpoint of $AC$,$\overrightarrow{e} = \frac{\overrightarrow{a} + \overrightarrow{c}}{2}$.
Since $F$ is the midpoint of $BD$,$\overrightarrow{f} = \frac{\overrightarrow{b} + \overrightarrow{d}}{2}$.
The given expression is $(\overrightarrow{AB} - \overrightarrow{BC}) + (\overrightarrow{AD} - \overrightarrow{DC}) = k \overrightarrow{FE}$.
Substituting the vectors: $(\overrightarrow{b} - \overrightarrow{a} - (\overrightarrow{c} - \overrightarrow{b})) + (\overrightarrow{d} - \overrightarrow{a} - (\overrightarrow{c} - \overrightarrow{d})) = k \overrightarrow{FE}$.
Simplifying: $(\overrightarrow{b} - \overrightarrow{a} - \overrightarrow{c} + \overrightarrow{b}) + (\overrightarrow{d} - \overrightarrow{a} - \overrightarrow{c} + \overrightarrow{d}) = k \overrightarrow{FE}$.
$(2\overrightarrow{b} - 2\overrightarrow{a} - 2\overrightarrow{c} + 2\overrightarrow{d}) = k \overrightarrow{FE}$.
$2(\overrightarrow{b} + \overrightarrow{d}) - 2(\overrightarrow{a} + \overrightarrow{c}) = k \overrightarrow{FE}$.
Using $\overrightarrow{b} + \overrightarrow{d} = 2\overrightarrow{f}$ and $\overrightarrow{a} + \overrightarrow{c} = 2\overrightarrow{e}$:
$2(2\overrightarrow{f}) - 2(2\overrightarrow{e}) = k \overrightarrow{FE}$.
$4(\overrightarrow{f} - \overrightarrow{e}) = k \overrightarrow{FE}$.
Since $\overrightarrow{FE} = \overrightarrow{e} - \overrightarrow{f}$,we have $4(-(\overrightarrow{e} - \overrightarrow{f})) = k \overrightarrow{FE}$.
$-4 \overrightarrow{FE} = k \overrightarrow{FE}$.
Therefore,$k = -4$.

(B) The vertices are $A(2, 2, 1)$,$B(1, 2, 2)$,and $C(2, 1, 2)$.
Calculate the side lengths:
$AB = \sqrt{(1-2)^2 + (2-2)^2 + (2-1)^2} = \sqrt{1+0+1} = \sqrt{2}$
$BC = \sqrt{(2-1)^2 + (1-2)^2 + (2-2)^2} = \sqrt{1+1+0} = \sqrt{2}$
$CA = \sqrt{(2-2)^2 + (2-1)^2 + (1-2)^2} = \sqrt{0+1+1} = \sqrt{2}$
Since $AB = BC = CA = \sqrt{2}$,the triangle is equilateral.
For an equilateral triangle,the orthocenter $H$ coincides with the centroid $G$.
$G = \left(\frac{2+1+2}{3}, \frac{2+2+1}{3}, \frac{1+2+2}{3}\right) = \left(\frac{5}{3}, \frac{5}{3}, \frac{5}{3}\right)$.
In an equilateral triangle,the length of the perpendicular from the centroid to any side is the distance from the centroid to the midpoint of that side.
For side $AB$,the midpoint $D$ is $\left(\frac{2+1}{2}, \frac{2+2}{2}, \frac{1+2}{2}\right) = \left(\frac{3}{2}, 2, \frac{3}{2}\right)$.
$l_1 = \text{distance } GD = \sqrt{\left(\frac{5}{3}-\frac{3}{2}\right)^2 + \left(\frac{5}{3}-2\right)^2 + \left(\frac{5}{3}-\frac{3}{2}\right)^2}$
$l_1 = \sqrt{\left(\frac{1}{6}\right)^2 + \left(-\frac{1}{3}\right)^2 + \left(\frac{1}{6}\right)^2} = \sqrt{\frac{1}{36} + \frac{4}{36} + \frac{1}{36}} = \sqrt{\frac{6}{36}} = \sqrt{\frac{1}{6}}$.
Since the triangle is equilateral,$l_1 = l_2 = l_3 = \sqrt{\frac{1}{6}}$.
Therefore,$l_1^2 + l_2^2 + l_3^2 = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}$.

(B) Let $\vec{a} = 2 \hat{i} + 2 \hat{j} + \hat{k}$ and $\vec{b} = 2 \hat{i} + 4 \hat{j} + 4 \hat{k}$.
Lengths are $|\vec{a}| = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{4+4+1} = 3$ and $|\vec{b}| = \sqrt{2^2 + 4^2 + 4^2} = \sqrt{4+16+16} = 6$.
The angle bisector $OC$ divides $AB$ in the ratio $|\vec{a}| : |\vec{b}| = 3 : 6 = 1 : 2$.
Using the section formula,the position vector of $C$ is $\vec{c} = \frac{1(\vec{b}) + 2(\vec{a})}{1+2} = \frac{(2 \hat{i} + 4 \hat{j} + 4 \hat{k}) + 2(2 \hat{i} + 2 \hat{j} + \hat{k})}{3} = \frac{6 \hat{i} + 8 \hat{j} + 6 \hat{k}}{3} = 2 \hat{i} + \frac{8}{3} \hat{j} + 2 \hat{k}$.
The length $OC = |\vec{c}| = \sqrt{2^2 + (\frac{8}{3})^2 + 2^2} = \sqrt{4 + \frac{64}{9} + 4} = \sqrt{8 + \frac{64}{9}} = \sqrt{\frac{72+64}{9}} = \sqrt{\frac{136}{9}} = \frac{\sqrt{136}}{3} = \frac{2 \sqrt{34}}{3}$.

(B) Given $\vec{b} = 2\hat{i} + 3\hat{j} - 5\hat{k}$ and $\vec{c} = 3\hat{i} - \hat{j} + \lambda\hat{k}$.
Then $\vec{b} + \vec{c} = (2+3)\hat{i} + (3-1)\hat{j} + (-5+\lambda)\hat{k} = 5\hat{i} + 2\hat{j} + (\lambda-5)\hat{k}$.
Since $\vec{r}$ is a unit vector along $\vec{b} + \vec{c}$,we have $\vec{r} = \frac{\vec{b} + \vec{c}}{|\vec{b} + \vec{c}|}$.
Given $\vec{r} \cdot \vec{a} = 3$,so $\frac{(\vec{b} + \vec{c}) \cdot \vec{a}}{|\vec{b} + \vec{c}|} = 3$.
Calculate $(\vec{b} + \vec{c}) \cdot \vec{a} = (5\hat{i} + 2\hat{j} + (\lambda-5)\hat{k}) \cdot (\hat{i} + 2\hat{j} + 3\hat{k}) = 5(1) + 2(2) + 3(\lambda-5) = 5 + 4 + 3\lambda - 15 = 3\lambda - 6$.
Calculate $|\vec{b} + \vec{c}| = \sqrt{5^2 + 2^2 + (\lambda-5)^2} = \sqrt{25 + 4 + \lambda^2 - 10\lambda + 25} = \sqrt{\lambda^2 - 10\lambda + 54}$.
Substituting these into the equation: $\frac{3\lambda - 6}{\sqrt{\lambda^2 - 10\lambda + 54}} = 3$.
Dividing by $3$: $\frac{\lambda - 2}{\sqrt{\lambda^2 - 10\lambda + 54}} = 1$.
Squaring both sides: $(\lambda - 2)^2 = \lambda^2 - 10\lambda + 54$.
$\lambda^2 - 4\lambda + 4 = \lambda^2 - 10\lambda + 54$.
$6\lambda = 50$.
$3\lambda = 25$.

(A) From the figure,the vector $\vec{R}$ is given by $\vec{R} = \vec{Q} - \vec{P}$.
Point $S$ lies on the line segment $PQ$ such that the distance $PS = b|\vec{R}|$.
This implies that the vector $\vec{PS} = b\vec{R} = b(\vec{Q} - \vec{P})$.
Using the triangle law of vector addition in $\triangle OPS$,we have $\vec{S} = \vec{P} + \vec{PS}$.
Substituting the expression for $\vec{PS}$,we get $\vec{S} = \vec{P} + b(\vec{Q} - \vec{P})$.
Simplifying this,we obtain $\vec{S} = \vec{P} + b\vec{Q} - b\vec{P} = (1-b)\vec{P} + b\vec{Q}$.

(D) Given that $\vec{s} = 4\vec{p} + 3\vec{q} + 5\vec{r}$.
Also,$\vec{s} = x(-\vec{p} + \vec{q} + \vec{r}) + y(\vec{p} - \vec{q} + \vec{r}) + z(-\vec{p} - \vec{q} + \vec{r})$.
Expanding the right side,we get $\vec{s} = (-x + y - z)\vec{p} + (x - y - z)\vec{q} + (x + y + z)\vec{r}$.
Since $\vec{p}, \vec{q}, \vec{r}$ are non-coplanar,they are linearly independent. Comparing coefficients:
$-x + y - z = 4$ $(1)$
$x - y - z = 3$ $(2)$
$x + y + z = 5$ $(3)$
Adding $(2)$ and $(3)$,we get $2x = 8 \Rightarrow x = 4$.
Substituting $x = 4$ in $(3)$,$y + z = 1$.
Substituting $x = 4$ in $(2)$,$4 - y - z = 3 \Rightarrow y + z = 1$ (consistent).
Substituting $x = 4$ in $(1)$,$-4 + y - z = 4 \Rightarrow y - z = 8$.
Solving $y + z = 1$ and $y - z = 8$,we get $2y = 9 \Rightarrow y = 4.5$ and $2z = -7 \Rightarrow z = -3.5$.
Finally,$2x + y + z = 2(4) + 4.5 - 3.5 = 8 + 1 = 9$.

(A) Let the position vectors of vertices $A, B, C$ be $\vec{a}, \vec{b}, \vec{c}$ respectively. Without loss of generality,let $\vec{a} = \vec{0}$.
Given $\vec{AB} = \vec{b} - \vec{a} = 2\hat{i}-\hat{j}+\hat{k}$,so $\vec{b} = 2\hat{i}-\hat{j}+\hat{k}$.
Given $\vec{CA} = \vec{a} - \vec{c} = \hat{i}-3\hat{j}-5\hat{k}$,so $\vec{c} = -(\hat{i}-3\hat{j}-5\hat{k}) = -\hat{i}+3\hat{j}+5\hat{k}$.
The centroid $G$ has position vector $\vec{g} = \frac{\vec{a}+\vec{b}+\vec{c}}{3} = \frac{\vec{0} + (2\hat{i}-\hat{j}+\hat{k}) + (-\hat{i}+3\hat{j}+5\hat{k})}{3} = \frac{\hat{i}+2\hat{j}+6\hat{k}}{3}$.
Now,$\overrightarrow{AG} = \vec{g} - \vec{a} = \frac{1}{3}(\hat{i}+2\hat{j}+6\hat{k})$. Thus,$|\overrightarrow{AG}|^2 = \frac{1}{9}(1^2+2^2+6^2) = \frac{41}{9}$.
$\overrightarrow{BG} = \vec{g} - \vec{b} = (\frac{1}{3}-2)\hat{i} + (\frac{2}{3}+1)\hat{j} + (2-1)\hat{k} = -\frac{5}{3}\hat{i} + \frac{5}{3}\hat{j} + \hat{k}$. Thus,$|\overrightarrow{BG}|^2 = \frac{25}{9} + \frac{25}{9} + 1 = \frac{59}{9}$.
$\overrightarrow{CG} = \vec{g} - \vec{c} = (\frac{1}{3}+1)\hat{i} + (\frac{2}{3}-3)\hat{j} + (2-5)\hat{k} = \frac{4}{3}\hat{i} - \frac{7}{3}\hat{j} - 3\hat{k}$. Thus,$|\overrightarrow{CG}|^2 = \frac{16}{9} + \frac{49}{9} + 9 = \frac{65+81}{9} = \frac{146}{9}$.
Finally,$6(|\overrightarrow{AG}|^2+|\overrightarrow{BG}|^2+|\overrightarrow{CG}|^2) = 6(\frac{41+59+146}{9}) = 6(\frac{246}{9}) = 6 \times \frac{82}{3} = 2 \times 82 = 164$.

(B) Let the two vectors be $\vec{a} = (1, -\sqrt{2})$ and $\vec{b} = (2, \sqrt{2})$.
Their sum is $\vec{s} = \vec{a} + \vec{b} = (1 + 2, -\sqrt{2} + \sqrt{2}) = (3, 0)$.
The magnitude of the sum vector $\vec{s}$ is given by $|\vec{s}| = \sqrt{3^2 + 0^2} = \sqrt{9} = 3$.

(C) The position vector of point $A$ is $\vec{a} = (1, 1, 0)$.
The position vector of point $B$ is $\vec{b} = (0, 1, 1)$.
The vector $\overrightarrow{AB}$ is given by the formula $\overrightarrow{AB} = \vec{b} - \vec{a}$.
Substituting the values,we get $\overrightarrow{AB} = (0 - 1, 1 - 1, 1 - 0)$.
Therefore,$\overrightarrow{AB} = (-1, 0, 1)$.

(C) Two vectors $\vec{a} = (a_1, a_2, a_3)$ and $\vec{b} = (b_1, b_2, b_3)$ have the same direction if $\vec{b} = k\vec{a}$ for some scalar $k > 0$.
Here,$\vec{a} = (1, 1, 2)$ and $\vec{b} = (2, 1, 0)$.
Checking for proportionality: $\frac{2}{1} \neq \frac{1}{1} \neq \frac{0}{2}$.
Since the ratios of the corresponding components are not equal,the vectors are not parallel.
Therefore,the directions of the two vectors are different.

(A) The given expression is $ < 2, 2, 2>$.
We evaluate the options to find an equivalent expression.
For option $A$: $- < -2, -2, -2>$.
By distributing the negative sign inside the vector,we get $-(-2), -(-2), -(-2) = < 2, 2, 2>$.
This matches the given expression.
Therefore,the correct option is $A$.

(B) The magnitude of a vector $\vec{a} = \langle x, y, z \rangle$ is given by the formula $|\vec{a}| = \sqrt{x^2 + y^2 + z^2}$.
Given $\vec{a} = \langle \frac{1}{3}, \frac{1}{3}, \frac{1}{3} \rangle$,we substitute the components into the formula:
$|\vec{a}| = \sqrt{(\frac{1}{3})^2 + (\frac{1}{3})^2 + (\frac{1}{3})^2}$
$|\vec{a}| = \sqrt{\frac{1}{9} + \frac{1}{9} + \frac{1}{9}}$
$|\vec{a}| = \sqrt{\frac{3}{9}}$
$|\vec{a}| = \sqrt{\frac{1}{3}}$
$|\vec{a}| = \frac{1}{\sqrt{3}}$
Thus,the correct option is $B$.

(A) Let $\vec{a} = (2, 2, -1)$.
The magnitude of the vector $\vec{a}$ is given by $|\vec{a}| = \sqrt{2^2 + 2^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.
The unit vector in the direction of $\vec{a}$ is denoted by $\hat{a}$ and is calculated as $\hat{a} = \frac{\vec{a}}{|\vec{a}|}$.
Substituting the values,we get $\hat{a} = \frac{1}{3}(2, 2, -1) = \left(\frac{2}{3}, \frac{2}{3}, \frac{-1}{3}\right)$.

(D) unit vector in the direction of a vector $\vec{a}$ is given by $\hat{a} = \frac{\vec{a}}{|\vec{a}|}$.
Given the vector $\vec{a} = (1, 0, 0)$.
The magnitude of the vector is $|\vec{a}| = \sqrt{1^2 + 0^2 + 0^2} = \sqrt{1} = 1$.
Therefore,the unit vector is $\hat{a} = \frac{(1, 0, 0)}{1} = (1, 0, 0)$.

(A) The vector $\overrightarrow{AB}$ is given by $(2-1)\hat{i} + (3-2)\hat{j} + (2-1)\hat{k} = 1\hat{i} + 1\hat{j} + 1\hat{k}$.
The vector $\overrightarrow{CD}$ is given by $(3-2)\hat{i} + (2-1)\hat{j} + (4-3)\hat{k} = 1\hat{i} + 1\hat{j} + 1\hat{k}$.
Since $\overrightarrow{AB} = \overrightarrow{CD}$,the vectors have the same direction.

(A) zero vector is a vector whose magnitude is $0$.
By definition,a zero vector has an indeterminate or arbitrary direction,which is often described as having no specific direction.
Therefore,the correct statement is that it has no direction.

(C) The coordinates of points are $P(2, 3, 1)$ and $Q(7, 15, 1)$.
The vector $\overrightarrow{PQ}$ is given by $(7 - 2)\hat{i} + (15 - 3)\hat{j} + (1 - 1)\hat{k}$.
$\overrightarrow{PQ} = 5\hat{i} + 12\hat{j} + 0\hat{k}$.
The magnitude $|\overrightarrow{PQ}|$ is calculated as $\sqrt{5^2 + 12^2 + 0^2}$.
$|\overrightarrow{PQ}| = \sqrt{25 + 144 + 0} = \sqrt{169} = 13$.

(C) Let the given vector be $\vec{a} = 3\hat{i} + 6\hat{j} + 2\hat{k}$.
First,find the magnitude of $\vec{a}$:
$|\vec{a}| = \sqrt{3^2 + 6^2 + 2^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7$.
Now,find the unit vector in the direction of $\vec{a}$:
$\hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{3}{7}\hat{i} + \frac{6}{7}\hat{j} + \frac{2}{7}\hat{k}$.
$A$ vector with magnitude $4$ in the direction of $\vec{a}$ is given by $4 \times \hat{a}$:
$4 \times \left(\frac{3}{7}\hat{i} + \frac{6}{7}\hat{j} + \frac{2}{7}\hat{k}\right) = \left(\frac{12}{7}, \frac{24}{7}, \frac{8}{7}\right)$.

(A) Let the given vector be $\vec{a} = (2, -2, 1)$.
First,find the magnitude of $\vec{a}$:
$|\vec{a}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.
The unit vector in the direction of $\vec{a}$ is $\hat{a} = \frac{\vec{a}}{|\vec{a}|} = \left(\frac{2}{3}, \frac{-2}{3}, \frac{1}{3}\right)$.
The unit vector in the opposite direction of $\vec{a}$ is $-\hat{a} = -\left(\frac{2}{3}, \frac{-2}{3}, \frac{1}{3}\right) = \left(\frac{-2}{3}, \frac{2}{3}, \frac{-1}{3}\right)$.

(C) Let $\bar{u} = \frac{\bar{x}}{|\bar{x}|}$ be the unit vector in the direction of $\bar{x}$.
Then the given expression is $-k \bar{u}$.
Since $k > 0$,the magnitude of $-k \bar{u}$ is $|-k| |\bar{u}| = k \times 1 = k$.
The negative sign indicates that the vector is in the opposite direction of $\bar{u}$,which is the opposite direction of $\bar{x}$.
Therefore,the vector is in the opposite direction of $\bar{x}$ and has a magnitude of $k$.

(A) Let the position vectors of points $P, Q, R, S, E,$ and $F$ be $\vec{p}, \vec{q}, \vec{r}, \vec{s}, \vec{e},$ and $\vec{f}$ respectively.
Given the equation: $\overrightarrow{SP} + 5\overrightarrow{SQ} + 6\overrightarrow{SR} = \overrightarrow{0}$.
Substituting the position vectors: $(\vec{p} - \vec{s}) + 5(\vec{q} - \vec{s}) + 6(\vec{r} - \vec{s}) = \overrightarrow{0}$.
$\vec{p} + 5\vec{q} + 6\vec{r} - 12\vec{s} = \overrightarrow{0} \Rightarrow \vec{s} = \frac{\vec{p} + 5\vec{q} + 6\vec{r}}{12}$.
Since $E$ is the mid-point of $PR$, $\vec{e} = \frac{\vec{p} + \vec{r}}{2}$.
Since $F$ is the mid-point of $QR$, $\vec{f} = \frac{\vec{q} + \vec{r}}{2}$.
Vector $\overrightarrow{EF} = \vec{f} - \vec{e} = \frac{\vec{q} + \vec{r}}{2} - \frac{\vec{p} + \vec{r}}{2} = \frac{\vec{q} - \vec{p}}{2}$.
Vector $\overrightarrow{ES} = \vec{s} - \vec{e} = \frac{\vec{p} + 5\vec{q} + 6\vec{r}}{12} - \frac{\vec{p} + \vec{r}}{2} = \frac{\vec{p} + 5\vec{q} + 6\vec{r} - 6\vec{p} - 6\vec{r}}{12} = \frac{5\vec{q} - 5\vec{p}}{12} = \frac{5}{12}(\vec{q} - \vec{p})$.
Therefore, the ratio of the lengths is $\frac{|\overrightarrow{EF}|}{|\overrightarrow{ES}|} = \frac{|\frac{1}{2}(\vec{q} - \vec{p})|}{|\frac{5}{12}(\vec{q} - \vec{p})|} = \frac{1/2}{5/12} = \frac{1}{2} \times \frac{12}{5} = \frac{6}{5} = 1.2$.

(B) Let the position vectors of $A, B, C, D, E, F$ be $\overline{a}, \overline{b}, \overline{c}, \overline{d}, \overline{e}, \overline{f}$ respectively.
$\therefore \overline{d} = \frac{\overline{b} + \overline{c}}{2}, \overline{e} = \frac{\overline{c} + \overline{a}}{2}, \overline{f} = \frac{\overline{a} + \overline{b}}{2}$.
Now,$\overline{AD} + \frac{2}{3} \overline{BE} + \frac{1}{3} \overline{CF} = (\overline{d} - \overline{a}) + \frac{2}{3}(\overline{e} - \overline{b}) + \frac{1}{3}(\overline{f} - \overline{c})$.
Substituting the values:
$= \frac{\overline{b} + \overline{c}}{2} - \overline{a} + \frac{2}{3}\left(\frac{\overline{c} + \overline{a}}{2} - \overline{b}\right) + \frac{1}{3}\left(\frac{\overline{a} + \overline{b}}{2} - \overline{c}\right)$.
$= \frac{\overline{b} + \overline{c} - 2\overline{a}}{2} + \frac{\overline{c} + \overline{a} - 2\overline{b}}{3} + \frac{\overline{a} + \overline{b} - 2\overline{c}}{6}$.
$= \frac{3(\overline{b} + \overline{c} - 2\overline{a}) + 2(\overline{c} + \overline{a} - 2\overline{b}) + (\overline{a} + \overline{b} - 2\overline{c})}{6}$.
$= \frac{3\overline{b} + 3\overline{c} - 6\overline{a} + 2\overline{c} + 2\overline{a} - 4\overline{b} + \overline{a} + \overline{b} - 2\overline{c}}{6}$.
$= \frac{-3\overline{a} + 3\overline{c}}{6} = \frac{3(\overline{c} - \overline{a})}{6} = \frac{1}{2}(\overline{c} - \overline{a}) = \frac{1}{2} \overline{AC}$.

(D) Given that $c = (x-2)a + b$ and $d = (2x+1)a - b$ are collinear vectors.
Since $a$ and $b$ are non-collinear,we can write $c = \lambda d$ for some scalar $\lambda$.
$(x-2)a + b = \lambda((2x+1)a - b)$
$(x-2)a + b = \lambda(2x+1)a - \lambda b$
Comparing the coefficients of $a$ and $b$,we get:
$x-2 = \lambda(2x+1)$ and $1 = -\lambda$.
From the second equation,$\lambda = -1$.
Substituting $\lambda = -1$ into the first equation:
$x-2 = -1(2x+1)$
$x-2 = -2x-1$
$3x = 1$
$x = \frac{1}{3}$

(B) Let the points be $\vec{p} = \hat{i} + 2\hat{j} - \hat{k}$ and $\vec{q} = -\hat{i} + \hat{j} + \hat{k}$.
For external division in the ratio $m:n = 1:2$,the position vector $\vec{r}$ is given by the formula:
$\vec{r} = \frac{m\vec{q} - n\vec{p}}{m - n}$
Substituting the values:
$\vec{r} = \frac{1(-\hat{i} + \hat{j} + \hat{k}) - 2(\hat{i} + 2\hat{j} - \hat{k})}{1 - 2}$
$\vec{r} = \frac{-\hat{i} + \hat{j} + \hat{k} - 2\hat{i} - 4\hat{j} + 2\hat{k}}{-1}$
$\vec{r} = \frac{-3\hat{i} - 3\hat{j} + 3\hat{k}}{-1}$
$\vec{r} = 3\hat{i} + 3\hat{j} - 3\hat{k}$

(A) Let $A$ be the origin $(0, 0, 0)$.
Then the coordinates of $B$ are $(3, 0, 4)$ and the coordinates of $C$ are $(5, -2, 4)$.
The median through $A$ meets the side $BC$ at its midpoint $M$.
The coordinates of the midpoint $M$ of $BC$ are given by $\left( \frac{3+5}{2}, \frac{0-2}{2}, \frac{4+4}{2} \right) = (4, -1, 4)$.
The length of the median $AM$ is the distance from $A(0, 0, 0)$ to $M(4, -1, 4)$.
Length $= \sqrt{(4-0)^2 + (-1-0)^2 + (4-0)^2} = \sqrt{16 + 1 + 16} = \sqrt{33} \text{ units}$.

(C) The position vectors are $\vec{A} = 2\hat{i}+\hat{j}-\hat{k}$,$\vec{B} = 3\hat{i}-2\hat{j}+\hat{k}$,and $\vec{C} = \hat{i}+4\hat{j}-3\hat{k}$.
$\overrightarrow{AB} = \vec{B} - \vec{A} = (3-2)\hat{i} + (-2-1)\hat{j} + (1-(-1))\hat{k} = \hat{i} - 3\hat{j} + 2\hat{k}$.
$|\overrightarrow{AB}| = \sqrt{1^2 + (-3)^2 + 2^2} = \sqrt{1+9+4} = \sqrt{14}$.
$\overrightarrow{BC} = \vec{C} - \vec{B} = (1-3)\hat{i} + (4-(-2))\hat{j} + (-3-1)\hat{k} = -2\hat{i} + 6\hat{j} - 4\hat{k}$.
$|\overrightarrow{BC}| = \sqrt{(-2)^2 + 6^2 + (-4)^2} = \sqrt{4+36+16} = \sqrt{56} = 2\sqrt{14}$.
$\overrightarrow{AC} = \vec{C} - \vec{A} = (1-2)\hat{i} + (4-1)\hat{j} + (-3-(-1))\hat{k} = -\hat{i} + 3\hat{j} - 2\hat{k}$.
$|\overrightarrow{AC}| = \sqrt{(-1)^2 + 3^2 + (-2)^2} = \sqrt{1+9+4} = \sqrt{14}$.
Since $\overrightarrow{AB} + \overrightarrow{AC} = (\hat{i} - 3\hat{j} + 2\hat{k}) + (-\hat{i} + 3\hat{j} - 2\hat{k}) = 0$,we have $\overrightarrow{AB} = -\overrightarrow{AC}$,which implies that the vectors are parallel and the points $A, B, C$ lie on the same line.
Thus,the points are collinear.

(D) Given $|\overline{u}|=2$. Let the direction angles be $\alpha, \beta, \gamma$.
We have $\alpha = 60^{\circ}$ and $\beta = 120^{\circ}$.
The direction cosines satisfy $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$.
Substituting the values: $(\cos 60^{\circ})^2 + (\cos 120^{\circ})^2 + \cos^2 \gamma = 1$.
$(\frac{1}{2})^2 + (-\frac{1}{2})^2 + \cos^2 \gamma = 1$.
$\frac{1}{4} + \frac{1}{4} + \cos^2 \gamma = 1 \Rightarrow \frac{1}{2} + \cos^2 \gamma = 1$.
$\cos^2 \gamma = \frac{1}{2} \Rightarrow \cos \gamma = \pm \frac{1}{\sqrt{2}}$.
The vector $\overline{u}$ is given by $|\overline{u}|(\cos \alpha \hat{i} + \cos \beta \hat{j} + \cos \gamma \hat{k})$.
$\overline{u} = 2(\frac{1}{2} \hat{i} - \frac{1}{2} \hat{j} \pm \frac{1}{\sqrt{2}} \hat{k})$.
$\overline{u} = \hat{i} - \hat{j} \pm \sqrt{2} \hat{k}$.
Wait,checking the options,the scalar factor $2$ is distributed. Let's re-evaluate: $\overline{u} = 2(\frac{1}{2} \hat{i} - \frac{1}{2} \hat{j} \pm \frac{1}{\sqrt{2}} \hat{k}) = \hat{i} - \hat{j} \pm \sqrt{2} \hat{k}$.
Given the options,the correct form is $2(\frac{1}{2}\hat{i} - \frac{1}{2}\hat{j} \pm \frac{1}{\sqrt{2}}\hat{k})$,which simplifies to $\hat{i}-\hat{j} \pm \sqrt{2}\hat{k}$. Option $D$ is the intended answer.

(A) Two vectors $\vec{a} = a_1 \hat{\imath} + a_2 \hat{\jmath} + a_3 \hat{k}$ and $\vec{b} = b_1 \hat{\imath} + b_2 \hat{\jmath} + b_3 \hat{k}$ are collinear if their corresponding components are proportional,i.e.,$\frac{a_1}{b_1} = \frac{a_2}{b_2} = \frac{a_3}{b_3}$.
Given vectors are $(2, -q, 3)$ and $(4, -5, 6)$.
Setting the ratios equal,we get:
$\frac{2}{4} = \frac{-q}{-5} = \frac{3}{6}$
Simplifying the fractions:
$\frac{1}{2} = \frac{q}{5} = \frac{1}{2}$
From $\frac{1}{2} = \frac{q}{5}$,we find:
$q = \frac{5}{2}$

(C) The magnitude of a vector remains invariant under the rotation of the coordinate system about the origin.
Therefore,the magnitude of $\vec{a}$ before rotation is equal to the magnitude of $\vec{a}$ after rotation.
$|\vec{a}|^2 = (2p)^2 + 1^2 = (p+1)^2 + 1^2$
$4p^2 + 1 = p^2 + 2p + 1 + 1$
$4p^2 + 1 = p^2 + 2p + 2$
$3p^2 - 2p - 1 = 0$
Factoring the quadratic equation:
$3p^2 - 3p + p - 1 = 0$
$3p(p-1) + 1(p-1) = 0$
$(3p+1)(p-1) = 0$
Thus,$p=1$ or $p=-\frac{1}{3}$.

(C) We know that $|\vec{x}-\vec{y}|^2 = |\vec{x}|^2 + |\vec{y}|^2 - 2(\vec{x} \cdot \vec{y})$.
Expanding the given expression:
$S = |\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2$
$S = ( |\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a} \cdot \vec{b} ) + ( |\vec{b}|^2 + |\vec{c}|^2 - 2\vec{b} \cdot \vec{c} ) + ( |\vec{c}|^2 + |\vec{a}|^2 - 2\vec{c} \cdot \vec{a} )$
$S = 2(|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2) - 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a})$
Given $|\vec{a}|=3, |\vec{b}|=5, |\vec{c}|=7$,so $|\vec{a}|^2=9, |\vec{b}|^2=25, |\vec{c}|^2=49$.
$S = 2(9+25+49) - 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 166 - 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a})$.
We know that $|\vec{a}+\vec{b}+\vec{c}|^2 \ge 0$,which implies $|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \ge 0$.
So,$2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \ge -(9+25+49) = -83$.
Therefore,$S = 166 - 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \le 166 - (-83) = 249$.
Thus,the expression does not exceed $249$.

(A) Let $\hat{a} = \frac{\overline{a}}{|\overline{a}|}$ and $\hat{b} = \frac{\overline{b}}{|\overline{b}|}$ be the unit vectors along $\overline{OA}$ and $\overline{OB}$ respectively.
The angle bisector of $\angle AOB$ is along the direction of the sum of the unit vectors,which is $\hat{a} + \hat{b}$.
Given that the vector along the angle bisector is $x\hat{a} + y\hat{b}$,we have $x\hat{a} + y\hat{b} = k(\hat{a} + \hat{b})$ for some scalar $k \neq 0$.
Comparing the coefficients of $\hat{a}$ and $\hat{b}$,we get $x = k$ and $y = k$.
Therefore,$x = y$,which implies $x - y = 0$.

(B) Given that $\bar{a}, \bar{b}, \bar{c}$ are unit vectors,so $|\bar{a}| = |\bar{b}| = |\bar{c}| = 1$.
Expanding the given equation:
$|\bar{a}+\bar{b}|^2+|\bar{a}+\bar{c}|^2=8$
$(|\bar{a}|^2+|\bar{b}|^2+2\bar{a} \cdot \bar{b}) + (|\bar{a}|^2+|\bar{c}|^2+2\bar{a} \cdot \bar{c}) = 8$
Since $|\bar{a}|^2 = |\bar{b}|^2 = |\bar{c}|^2 = 1$,we have:
$(1+1+2\bar{a} \cdot \bar{b}) + (1+1+2\bar{a} \cdot \bar{c}) = 8$
$4 + 2(\bar{a} \cdot \bar{b} + \bar{a} \cdot \bar{c}) = 8$
$2(\bar{a} \cdot \bar{b} + \bar{a} \cdot \bar{c}) = 4$
$\bar{a} \cdot \bar{b} + \bar{a} \cdot \bar{c} = 2$.
Since the maximum value of the dot product of two unit vectors is $1$,the only way for the sum of two dot products to be $2$ is if $\bar{a} \cdot \bar{b} = 1$ and $\bar{a} \cdot \bar{c} = 1$.
This implies $\bar{a} = \bar{b} = \bar{c}$.
Now,calculate the required expression:
$|\bar{a}+3\bar{b}|^2+|\bar{a}+3\bar{c}|^2 = |\bar{a}+3\bar{a}|^2+|\bar{a}+3\bar{a}|^2$
$= |4\bar{a}|^2 + |4\bar{a}|^2$
$= 16|\bar{a}|^2 + 16|\bar{a}|^2$
$= 16(1) + 16(1) = 32$.

(C) Given equations are:
$(1)$ $\bar{c} = 5\bar{a} + 6\bar{b}$
$(2)$ $3\bar{c} = \bar{a} - 4\bar{b}$
Multiply equation $(1)$ by $3$:
$3\bar{c} = 15\bar{a} + 18\bar{b}$
Now,equate this with equation $(2)$:
$15\bar{a} + 18\bar{b} = \bar{a} - 4\bar{b}$
$14\bar{a} = -22\bar{b}$
$\bar{a} = -\frac{11}{7}\bar{b}$
Since $\bar{a}$ is a scalar multiple of $\bar{b}$ with a negative constant,$\bar{a}$ and $\bar{b}$ are collinear and in opposite directions.
Substitute $\bar{a} = -\frac{11}{7}\bar{b}$ into equation $(1)$:
$\bar{c} = 5(-\frac{11}{7}\bar{b}) + 6\bar{b} = -\frac{55}{7}\bar{b} + \frac{42}{7}\bar{b} = -\frac{13}{7}\bar{b}$
Since $\bar{c}$ is also a negative scalar multiple of $\bar{b}$,$\bar{c}$ and $\bar{b}$ are in opposite directions.
Since $\bar{a} = -\frac{11}{7}\bar{b}$ and $\bar{c} = -\frac{13}{7}\bar{b}$,we have $\bar{a} = \frac{11}{13}\bar{c}$.
Since $\bar{a}$ is a positive scalar multiple of $\bar{c}$,$\bar{a}$ and $\bar{c}$ are in the same direction.
Thus,$\bar{a}$ and $\bar{c}$ are in the same direction,while $\bar{a}$ and $\bar{b}$ are in opposite directions.

(C) Given that $\bar{s} = 4\bar{p} + 3\bar{q} + 5\bar{r}$.
Also,$\bar{s} = x(-\bar{p} + \bar{q} + \bar{r}) + y(\bar{p} - \bar{q} + \bar{r}) + z(-\bar{p} - \bar{q} + \bar{r})$.
Expanding the right side,we get $\bar{s} = (-x + y - z)\bar{p} + (x - y - z)\bar{q} + (x + y + z)\bar{r}$.
Comparing the coefficients of $\bar{p}, \bar{q}, \bar{r}$ with the given expression for $\bar{s}$,we have:
$-x + y - z = 4$ $(i)$
$x - y - z = 3$ $(ii)$
$x + y + z = 5$ $(iii)$
Adding $(ii)$ and $(iii)$,we get $2x = 8 \implies x = 4$.
Adding $(i)$ and $(ii)$,we get $-2z = 7 \implies z = -\frac{7}{2}$.
Substituting $x=4$ and $z=-\frac{7}{2}$ into $(iii)$,we get $4 + y - \frac{7}{2} = 5 \implies y = 1 + \frac{7}{2} = \frac{9}{2}$.
Now,calculate $2x + y + z = 2(4) + \frac{9}{2} - \frac{7}{2} = 8 + \frac{2}{2} = 8 + 1 = 9$.

(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) 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 $\overline{AD}$:
$|\overline{AD}| = \sqrt{4^2 + (-1)^2 + 4^2}$
$|\overline{AD}| = \sqrt{16 + 1 + 16}$
$|\overline{AD}| = \sqrt{33} \text{ units}$.

(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 $\frac{\vec{v}}{|\vec{v}|} = \frac{-\hat{i} + 2 \hat{j} - \hat{k}}{\sqrt{6}} = \frac{1}{\sqrt{6}}(-\hat{i} + 2 \hat{j} - \hat{k})$.

(A) Let $\vec{a}$ and $\vec{b}$ be the adjacent sides of a parallelogram,where $\vec{a} = 2 \hat{i}-4 \hat{j}+5 \hat{k}$ and $\vec{b} = \hat{i}-2 \hat{j}-3 \hat{k}$.
One diagonal of the parallelogram is given by $\vec{c} = \vec{a} + \vec{b}$.
$\vec{c} = (2+1) \hat{i} + (-4-2) \hat{j} + (5-3) \hat{k} = 3 \hat{i} - 6 \hat{j} + 2 \hat{k}$.
The magnitude of $\vec{c}$ is $|\vec{c}| = \sqrt{3^2 + (-6)^2 + 2^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7$.
The unit vector parallel to the diagonal $\vec{c}$ is $\hat{c} = \frac{\vec{c}}{|\vec{c}|} = \frac{3 \hat{i} - 6 \hat{j} + 2 \hat{k}}{7} = \frac{3}{7} \hat{i} - \frac{6}{7} \hat{j} + \frac{2}{7} \hat{k}$.

(A) Given: $\overline{a} = 4 \hat{i} + 13 \hat{j} - 18 \hat{k}$,$\overline{b} = \hat{i} - 2 \hat{j} + 3 \hat{k}$,and $\overline{c} = 2 \hat{i} + 3 \hat{j} - 4 \hat{k}$.
We are given the relation $\overline{a} = m \overline{b} + n \overline{c}$.
Substituting the vectors,we get:
$4 \hat{i} + 13 \hat{j} - 18 \hat{k} = m(\hat{i} - 2 \hat{j} + 3 \hat{k}) + n(2 \hat{i} + 3 \hat{j} - 4 \hat{k})$
$4 \hat{i} + 13 \hat{j} - 18 \hat{k} = (m + 2n) \hat{i} + (-2m + 3n) \hat{j} + (3m - 4n) \hat{k}$
Comparing the coefficients of $\hat{i}, \hat{j},$ and $\hat{k}$ on both sides,we obtain the following system of linear equations:
$1) \ m + 2n = 4$
$2) \ -2m + 3n = 13$
Multiply equation $(1)$ by $2$ to get $2m + 4n = 8$. Adding this to equation $(2)$:
$(2m + 4n) + (-2m + 3n) = 8 + 13$
$7n = 21 \Rightarrow n = 3$
Substitute $n = 3$ into equation $(1)$:
$m + 2(3) = 4 \Rightarrow m + 6 = 4 \Rightarrow m = -2$
Thus,$m + n = -2 + 3 = 1$.

(B) Given that $\bar{a}+2 \bar{b}$ is collinear with $\bar{c}$,there exists a non-zero scalar $n$ such that $\bar{a}+2 \bar{b} = n \bar{c}$. (Equation $1$)
Similarly,since $\bar{b}+3 \bar{c}$ is collinear with $\bar{a}$,there exists a non-zero scalar $m$ such that $\bar{b}+3 \bar{c} = m \bar{a}$. (Equation $2$)
From Equation $2$,we have $\bar{b} = m \bar{a} - 3 \bar{c}$.
Substitute this into Equation $1$: $\bar{a} + 2(m \bar{a} - 3 \bar{c}) = n \bar{c}$.
This simplifies to: $\bar{a} + 2m \bar{a} - 6 \bar{c} = n \bar{c}$.
$(1 + 2m) \bar{a} = (n + 6) \bar{c}$.
Since $\bar{a}$ and $\bar{c}$ are non-zero and not collinear,the coefficients must be zero.
Therefore,$1 + 2m = 0 \Rightarrow m = -\frac{1}{2}$ and $n + 6 = 0 \Rightarrow n = -6$.
Substituting $n = -6$ into Equation $1$,we get $\bar{a} + 2 \bar{b} = -6 \bar{c}$.

(C) The magnitude of a vector remains invariant under the rotation of the coordinate system about the origin.
Given the components of vector $\bar{a}$ in the original system are $(1, 2p)$,its squared magnitude is:
$|\bar{a}|^2 = 1^2 + (2p)^2 = 1 + 4p^2$
In the new system,the components are $(1, p+1)$,so its squared magnitude is:
$|\bar{b}|^2 = 1^2 + (p+1)^2 = 1 + p^2 + 2p + 1 = p^2 + 2p + 2$
Since $|\bar{a}|^2 = |\bar{b}|^2$,we have:
$1 + 4p^2 = p^2 + 2p + 2$
$3p^2 - 2p - 1 = 0$
Factoring the quadratic equation:
$(3p + 1)(p - 1) = 0$
Thus,$p = -\frac{1}{3}$ or $p = 1$.

(A) Given that $|\overline{a}|=2, |\overline{b}|=3, |\overline{c}|=5$ and the angle between each pair of vectors is $60^{\circ}$.
We know that $\overline{a} \cdot \overline{b} = |\overline{a}||\overline{b}| \cos 60^{\circ} = (2)(3)(\frac{1}{2}) = 3$.
Similarly,$\overline{b} \cdot \overline{c} = |\overline{b}||\overline{c}| \cos 60^{\circ} = (3)(5)(\frac{1}{2}) = \frac{15}{2}$.
And $\overline{c} \cdot \overline{a} = |\overline{c}||\overline{a}| \cos 60^{\circ} = (5)(2)(\frac{1}{2}) = 5$.
Now,$|\overline{a}+\overline{b}+\overline{c}|^2 = |\overline{a}|^2+|\overline{b}|^2+|\overline{c}|^2 + 2(\overline{a} \cdot \overline{b} + \overline{b} \cdot \overline{c} + \overline{c} \cdot \overline{a})$.
Substituting the values: $|\overline{a}+\overline{b}+\overline{c}|^2 = 2^2 + 3^2 + 5^2 + 2(3 + \frac{15}{2} + 5)$.
$|\overline{a}+\overline{b}+\overline{c}|^2 = 4 + 9 + 25 + 2(\frac{6+15+10}{2}) = 38 + 31 = 69$.
Therefore,$|\overline{a}+\overline{b}+\overline{c}| = \sqrt{69}$.