Basic , Modulus and Algebra of vectors Questions in English

(A) Let $\vec{a}$ and $\vec{b}$ be the adjacent sides of the parallelogram,and $\vec{d_1}$ and $\vec{d_2}$ be the diagonals.
We know that $\vec{d_1} = \vec{a} + \vec{b}$ and $\vec{d_2} = \vec{a} - \vec{b}$ (or vice versa).
Thus,$\vec{a} = \frac{\vec{d_1} + \vec{d_2}}{2} = \frac{(3 \hat{i} + 6 \hat{j} - 2 \hat{k}) + (-\hat{i} - 2 \hat{j} - 8 \hat{k})}{2} = \frac{2 \hat{i} + 4 \hat{j} - 10 \hat{k}}{2} = \hat{i} + 2 \hat{j} - 5 \hat{k}$.
The length of side $\vec{a}$ is $|\vec{a}| = \sqrt{1^2 + 2^2 + (-5)^2} = \sqrt{1 + 4 + 25} = \sqrt{30}$.
Similarly,$\vec{b} = \frac{\vec{d_1} - \vec{d_2}}{2} = \frac{(3 \hat{i} + 6 \hat{j} - 2 \hat{k}) - (-\hat{i} - 2 \hat{j} - 8 \hat{k})}{2} = \frac{4 \hat{i} + 8 \hat{j} + 6 \hat{k}}{2} = 2 \hat{i} + 4 \hat{j} + 3 \hat{k}$.
The length of side $\vec{b}$ is $|\vec{b}| = \sqrt{2^2 + 4^2 + 3^2} = \sqrt{4 + 16 + 9} = \sqrt{29}$.
Comparing the lengths $\sqrt{30}$ and $\sqrt{29}$,the shorter side is $\sqrt{29}$.

(D) Let $\vec{AB} = \hat{i}+\hat{j}+\hat{k}$ and $\vec{AC} = \hat{i}+3\hat{j}+5\hat{k}$.
Let $M$ be the midpoint of side $BC$. The median through $A$ is represented by the vector $\vec{AM}$.
In $\triangle ABC$,by the triangle law of vector addition,the position vector of the midpoint $M$ of $BC$ relative to $A$ is given by the average of the vectors $\vec{AB}$ and $\vec{AC}$:
$\vec{AM} = \frac{1}{2}(\vec{AB} + \vec{AC})$
Substituting the given vectors:
$\vec{AM} = \frac{1}{2}((\hat{i}+\hat{j}+\hat{k}) + (\hat{i}+3\hat{j}+5\hat{k}))$
$\vec{AM} = \frac{1}{2}(2\hat{i} + 4\hat{j} + 6\hat{k})$
$\vec{AM} = \hat{i} + 2\hat{j} + 3\hat{k}$
The length of the median is the magnitude of vector $\vec{AM}$:
$|\vec{AM}| = \sqrt{(1)^2 + (2)^2 + (3)^2}$
$|\vec{AM}| = \sqrt{1 + 4 + 9}$
$|\vec{AM}| = \sqrt{14}$

(B) Let the position vectors of vertices $A, B, C, D$ be $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ respectively.
Since $ABCD$ is a parallelogram,$\vec{d} = \vec{a} + \vec{c} - \vec{b}$.
$L$ is the mid-point of $BC$,so $\vec{l} = \frac{\vec{b} + \vec{c}}{2}$.
$M$ is the mid-point of $CD$,so $\vec{m} = \frac{\vec{c} + \vec{d}}{2} = \frac{\vec{c} + (\vec{a} + \vec{c} - \vec{b})}{2} = \frac{\vec{a} + 2\vec{c} - \vec{b}}{2}$.
Now,$\vec{AL} = \vec{l} - \vec{a} = \frac{\vec{b} + \vec{c}}{2} - \vec{a} = \frac{\vec{b} + \vec{c} - 2\vec{a}}{2}$.
And $\vec{AM} = \vec{m} - \vec{a} = \frac{\vec{a} + 2\vec{c} - \vec{b}}{2} - \vec{a} = \frac{2\vec{c} - \vec{b} - \vec{a}}{2}$.
Adding these,$\vec{AL} + \vec{AM} = \frac{\vec{b} + \vec{c} - 2\vec{a} + 2\vec{c} - \vec{b} - \vec{a}}{2} = \frac{3\vec{c} - 3\vec{a}}{2} = \frac{3}{2}(\vec{c} - \vec{a}) = \frac{3}{2} \vec{AC}$.
Thus,$AL + AM = \frac{3}{2} AC$.

(C) In $\triangle PQR$,$L, M, N$ are midpoints of $PQ, QR, RP$ respectively.
By the midpoint theorem,$\overrightarrow{LM} = \frac{1}{2} \overrightarrow{PR}$,$\overrightarrow{MN} = \frac{1}{2} \overrightarrow{PQ}$,and $\overrightarrow{NL} = \frac{1}{2} \overrightarrow{QR}$.
Also,$\overrightarrow{QM} = \frac{1}{2} \overrightarrow{QR} = \overrightarrow{NL}$,$\overrightarrow{LN} = \frac{1}{2} \overrightarrow{PR} = \overrightarrow{ML}$,$\overrightarrow{RN} = \frac{1}{2} \overrightarrow{RP} = -\frac{1}{2} \overrightarrow{PR} = -\overrightarrow{ML}$.
Substituting these into the expression:
$\overrightarrow{QM} + \overrightarrow{LN} + \overrightarrow{ML} + \overrightarrow{RN} - \overrightarrow{MN} - \overrightarrow{QL}$
$= \overrightarrow{NL} + \overrightarrow{ML} + \overrightarrow{ML} - \overrightarrow{ML} - \overrightarrow{MN} - \overrightarrow{QL}$
Since $\overrightarrow{QL} = -\overrightarrow{LQ} = -\overrightarrow{MN}$,the expression simplifies to $\vec{0}$.

(B) For an equilateral triangle,the orthocenter,circumcenter,centroid,and incenter all coincide at the same point.
Given that the origin is the orthocenter,it is also the centroid of the triangle.
The centroid of a triangle with vertices $\vec{a}, \vec{b}, \vec{c}$ is given by $\frac{\vec{a}+\vec{b}+\vec{c}}{3}$.
Since the centroid is the origin,we have:
$\frac{\vec{a}+\vec{b}+\vec{c}}{3} = \vec{0}$
$\vec{a}+\vec{b}+\vec{c} = \vec{0}$
Therefore,$\vec{a}+\vec{b} = -\vec{c}$.

(A) From the geometry of the quadrilateral,we have the vector relations:
$SQ = SP + PQ = (a-b) + a = 2a - b$
$SM = SQ + QM = (2a - b) + \frac{b}{2} = 2a - \frac{b}{2}$
$SM = \frac{1}{2}(4a - b)$
Comparing this with $SM = m(4a - b)$,we get $m = \frac{1}{2}$.
Given $SX = \frac{4}{5}SM$,we substitute the expression for $SM$:
$SX = \frac{4}{5} \cdot \frac{1}{2}(4a - b) = \frac{2}{5}(4a - b)$
Comparing this with $SX = n(4a - b)$,we get $n = \frac{2}{5}$.
Therefore,$m + n = \frac{1}{2} + \frac{2}{5} = \frac{5+4}{10} = \frac{9}{10}$.

(C) In $\triangle PAC$,by triangle law of vector addition,$\vec{PA} + \vec{AC} = \vec{PC}$ ...$(i)$
In $\triangle PBC$,by triangle law of vector addition,$\vec{PB} + \vec{BC} = \vec{PC}$ ...$(ii)$
Adding equations $(i)$ and $(ii)$:
$\vec{PA} + \vec{PB} + \vec{AC} + \vec{BC} = 2\vec{PC}$
Since $C$ is the mid-point of $AB$,we have $\vec{AC} = -\vec{BC}$,or $\vec{AC} + \vec{BC} = 0$.
Therefore,$\vec{PA} + \vec{PB} = 2\vec{PC}$.

(A) Let the position vectors of vertices $A, B, C, D$ be $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ respectively.
Let $P, Q, R, S$ be the mid-points of sides $AB, BC, CD, DA$ respectively.
Then $P = \frac{\vec{a}+\vec{b}}{2}$,$Q = \frac{\vec{b}+\vec{c}}{2}$,$R = \frac{\vec{c}+\vec{d}}{2}$,$S = \frac{\vec{d}+\vec{a}}{2}$.
The mid-point of $PR$ is $\frac{P+R}{2} = \frac{\frac{\vec{a}+\vec{b}}{2} + \frac{\vec{c}+\vec{d}}{2}}{2} = \frac{\vec{a}+\vec{b}+\vec{c}+\vec{d}}{4}$.
The mid-point of $SQ$ is $\frac{S+Q}{2} = \frac{\frac{\vec{d}+\vec{a}}{2} + \frac{\vec{b}+\vec{c}}{2}}{2} = \frac{\vec{a}+\vec{b}+\vec{c}+\vec{d}}{4}$.
Since the mid-points of $PR$ and $SQ$ coincide,let this point be $E$.
Thus,the position vector of $E$ is $\vec{e} = \frac{\vec{a}+\vec{b}+\vec{c}+\vec{d}}{4}$.
This implies $4\vec{e} = \vec{a}+\vec{b}+\vec{c}+\vec{d}$.
For any point $O$ with position vector $\vec{o}$,we have $\vec{OA} = \vec{a}-\vec{o}$,$\vec{OB} = \vec{b}-\vec{o}$,$\vec{OC} = \vec{c}-\vec{o}$,$\vec{OD} = \vec{d}-\vec{o}$,and $\vec{OE} = \vec{e}-\vec{o}$.
Then $\vec{OA} + \vec{OB} + \vec{OC} + \vec{OD} = (\vec{a}+\vec{b}+\vec{c}+\vec{d}) - 4\vec{o} = 4\vec{e} - 4\vec{o} = 4(\vec{e}-\vec{o}) = 4\vec{OE}$.
Comparing this with $x\vec{OE}$,we get $x = 4$.

(A) In a parallelogram $ABCD$,by the triangle law of vector addition:
$\vec{AC} = \vec{AB} + \vec{BC} \quad \dots(i)$
$\vec{BD} = \vec{BA} + \vec{AD} = -\vec{AB} + \vec{AD} \quad \dots(ii)$
Since $ABCD$ is a parallelogram,$\vec{AD} = \vec{BC}$.
Subtracting equation $(ii)$ from $(i)$:
$\vec{AC} - \vec{BD} = (\vec{AB} + \vec{BC}) - (-\vec{AB} + \vec{AD})$
$\vec{AC} - \vec{BD} = \vec{AB} + \vec{BC} + \vec{AB} - \vec{BC}$
$\vec{AC} - \vec{BD} = 2\vec{AB}$

(C) In a triangle $ABC$,let the position vectors of vertices $A, B, C$ be $\vec{a}, \vec{b}, \vec{c}$ respectively.
Given $\vec{u} = \vec{AB} = \vec{b} - \vec{a}$ and $\vec{v} = \vec{AC} = \vec{c} - \vec{a}$.
Since $D$ is the midpoint of $BC$,its position vector is $\vec{d} = \frac{\vec{b} + \vec{c}}{2}$.
Then,$\vec{AD} = \vec{d} - \vec{a} = \frac{\vec{b} + \vec{c}}{2} - \vec{a} = \frac{\vec{b} - \vec{a} + \vec{c} - \vec{a}}{2} = \frac{\vec{u} + \vec{v}}{2}$.

(A) Let the position vectors of vertices $A, B, C$ be $\vec{a}, \vec{b}, \vec{c}$ respectively.
Since $G$ is the centroid of $\triangle ABC$,its position vector $\vec{g}$ is given by $\vec{g} = \frac{\vec{a} + \vec{b} + \vec{c}}{3}$.
This implies $\vec{a} + \vec{b} + \vec{c} = 3\vec{g}$.
Now,the sum of the vectors $\vec{GA} + \vec{GB} + \vec{GC}$ is:
$\vec{GA} + \vec{GB} + \vec{GC} = (\vec{a} - \vec{g}) + (\vec{b} - \vec{g}) + (\vec{c} - \vec{g})$
$= (\vec{a} + \vec{b} + \vec{c}) - 3\vec{g}$
$= 3\vec{g} - 3\vec{g} = \vec{0}$.

(B) Given position vectors are $\vec{a} = 2\hat{i}-\hat{j}-\hat{k}$,$\vec{b} = 5\hat{i}+\hat{j}-2\hat{k}$,and $\vec{c} = -13\hat{i}-11\hat{j}+4\hat{k}$.
First,calculate the vectors $\vec{AB}$,$\vec{BC}$,$\vec{AC}$,and $\vec{CB}$:
$\vec{AB} = \vec{b} - \vec{a} = (5-2)\hat{i} + (1-(-1))\hat{j} + (-2-(-1))\hat{k} = 3\hat{i} + 2\hat{j} - \hat{k}$.
$\vec{BC} = \vec{c} - \vec{b} = (-13-5)\hat{i} + (-11-1)\hat{j} + (4-(-2))\hat{k} = -18\hat{i} - 12\hat{j} + 6\hat{k}$.
$\vec{AC} = \vec{c} - \vec{a} = (-13-2)\hat{i} + (-11-(-1))\hat{j} + (4-(-1))\hat{k} = -15\hat{i} - 10\hat{j} + 5\hat{k}$.
$\vec{CB} = \vec{b} - \vec{c} = (5-(-13))\hat{i} + (1-(-11))\hat{j} + (-2-4)\hat{k} = 18\hat{i} + 12\hat{j} - 6\hat{k}$.
Now,solve for $\lambda$ in $\vec{AB} = \lambda \vec{BC}$:
$3\hat{i} + 2\hat{j} - \hat{k} = \lambda (-18\hat{i} - 12\hat{j} + 6\hat{k}) = -6\lambda (3\hat{i} + 2\hat{j} - \hat{k})$.
Thus,$-6\lambda = 1$,which gives $\lambda = -\frac{1}{6}$.
Next,solve for $\mu$ in $\vec{AC} = \mu \vec{CB}$:
$-15\hat{i} - 10\hat{j} + 5\hat{k} = \mu (18\hat{i} + 12\hat{j} - 6\hat{k})$.
Dividing the components: $-15 = 18\mu \implies \mu = -\frac{15}{18} = -\frac{5}{6}$.
Finally,$\lambda + \mu = -\frac{1}{6} + (-\frac{5}{6}) = -\frac{6}{6} = -1$.

(B) Given that $\bar{a}$ and $\bar{b}$ are position vectors of $A$ and $B$.
Thus,$\overline{AB} = \bar{b} - \bar{a}$.
Given $\overline{AC} = 3 \overline{AB} = 3(\bar{b} - \bar{a})$.
Since $\overline{AC} = \vec{c} - \vec{a}$,we have $\vec{c} - \vec{a} = 3\bar{b} - 3\bar{a}$,which implies $\vec{c} = 3\bar{b} - 2\bar{a}$.
Given $\overline{BD} = 2 \overline{BA} = 2(\bar{a} - \bar{b})$.
Since $\overline{BD} = \vec{d} - \vec{b}$,we have $\vec{d} - \vec{b} = 2\bar{a} - 2\bar{b}$,which implies $\vec{d} = 2\bar{a} - \bar{b}$.
Now,$\overline{CD} = \vec{d} - \vec{c} = (2\bar{a} - \bar{b}) - (3\bar{b} - 2\bar{a}) = 2\bar{a} - \bar{b} - 3\bar{b} + 2\bar{a} = 4\bar{a} - 4\bar{b}$.

(B) Let $\bar{u} = \bar{i} - 7\bar{j} + 2\bar{k}$ be the vector along the internal bisector. The magnitude $|\bar{u}| = \sqrt{1^2 + (-7)^2 + 2^2} = \sqrt{1 + 49 + 4} = \sqrt{54} = 3\sqrt{6}$.
Let $\bar{b} = -2\bar{i} - \bar{j} + 2\bar{k}$. The magnitude $|\bar{b}| = \sqrt{(-2)^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = 3$.
The unit vector along $\bar{b}$ is $\hat{b} = \frac{-2\bar{i} - \bar{j} + 2\bar{k}}{3} = -\frac{2}{3}\bar{i} - \frac{1}{3}\bar{j} + \frac{2}{3}\bar{k}$.
Let $\hat{a} = x\bar{i} + y\bar{j} + z\bar{k}$ be the unit vector along $\bar{a}$.
The internal bisector is along $\hat{a} + \hat{b}$. Thus,$\hat{a} + \hat{b} = k\bar{u}$ for some scalar $k > 0$.
$\hat{a} = k(\bar{i} - 7\bar{j} + 2\bar{k}) - (-\frac{2}{3}\bar{i} - \frac{1}{3}\bar{j} + \frac{2}{3}\bar{k}) = (k + \frac{2}{3})\bar{i} + (-7k + \frac{1}{3})\bar{j} + (2k - \frac{2}{3})\bar{k}$.
Since $|\hat{a}| = 1$,we have $(k + \frac{2}{3})^2 + (-7k + \frac{1}{3})^2 + (2k - \frac{2}{3})^2 = 1$.
$(k^2 + \frac{4}{3}k + \frac{4}{9}) + (49k^2 - \frac{14}{3}k + \frac{1}{9}) + (4k^2 - \frac{8}{3}k + \frac{4}{9}) = 1$.
$54k^2 - 6k + 1 = 1 \implies 54k^2 - 6k = 0 \implies 6k(9k - 1) = 0$.
Since $k > 0$,$k = \frac{1}{9}$.
Then $x = k + \frac{2}{3} = \frac{1}{9} + \frac{6}{9} = \frac{7}{9}$.

(A) Given vectors are $\bar{a} = 2\bar{i} - 3\bar{j} + 5\bar{k}$ and $\bar{b} = -\bar{i} + 3\bar{j} + 3\bar{k}$.
First,calculate the vector $\bar{c} = \bar{a} - \bar{b}$:
$\bar{c} = (2 - (-1))\bar{i} + (-3 - 3)\bar{j} + (5 - 3)\bar{k} = 3\bar{i} - 6\bar{j} + 2\bar{k}$.
Next,find the magnitude of $\bar{c}$:
$|\bar{c}| = \sqrt{3^2 + (-6)^2 + 2^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7$.
The unit vector in the direction of $\bar{c}$ is $\hat{c} = \frac{\bar{c}}{|\bar{c}|} = \frac{3\bar{i} - 6\bar{j} + 2\bar{k}}{7}$.
The required vector of magnitude $28$ units is $28 \times \hat{c}$:
$28 \times \left( \frac{3\bar{i} - 6\bar{j} + 2\bar{k}}{7} \right) = 4(3\bar{i} - 6\bar{j} + 2\bar{k}) = 12\bar{i} - 24\bar{j} + 8\bar{k}$.

(B) Let the position vectors be $\vec{a} = \overline{i}-2 \overline{j}+\overline{k}$,$\vec{b} = \overline{i}+\overline{j}-2 \overline{k}$,$\vec{c} = 2 \overline{i}-\overline{j}-\overline{k}$,and $\vec{d} = \overline{i}+\overline{j}+\overline{k}$.
Point $P$ divides $AB$ in ratio $2:1$ internally: $\vec{p} = \frac{2\vec{b} + 1\vec{a}}{2+1} = \frac{2(\overline{i}+\overline{j}-2 \overline{k}) + (\overline{i}-2 \overline{j}+\overline{k})}{3} = \frac{3\overline{i} - 3\overline{k}}{3} = \overline{i} - \overline{k}$.
Point $Q$ divides $CD$ in ratio $1:2$ externally: $\vec{q} = \frac{1\vec{d} - 2\vec{c}}{1-2} = \frac{(\overline{i}+\overline{j}+\overline{k}) - 2(2 \overline{i}-\overline{j}-\overline{k})}{-1} = \frac{-3\overline{i} + 3\overline{j} + 3\overline{k}}{-1} = 3\overline{i} - 3\overline{j} - 3\overline{k}$.
Let the point $R$ with position vector $\vec{r} = 5\overline{i}-6\overline{j}-5\overline{k}$ divide $PQ$ in ratio $k:1$.
Then $\vec{r} = \frac{k\vec{q} + 1\vec{p}}{k+1} \implies (k+1)(5\overline{i}-6\overline{j}-5\overline{k}) = k(3\overline{i}-3\overline{j}-3\overline{k}) + (\overline{i}-\overline{k})$.
Comparing components:
$x$-component: $5k+5 = 3k+1 \implies 2k = -4 \implies k = -2$.
Thus,the ratio is $-2:1$.

(A) In a regular hexagon $ABCDEF$,the center $O$ is such that $\overrightarrow{OA}=\overrightarrow{BC}=\vec{b}$ and $\overrightarrow{AB}=\overrightarrow{OC}=\vec{a}$.
Also,$\overrightarrow{CD}=\overrightarrow{AF}$ and $\overrightarrow{DE}=\overrightarrow{BA}=-\vec{a}$.
By the polygon law of vector addition,the sum of vectors along the sides of a closed polygon is zero:
$\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}+\overrightarrow{DE}+\overrightarrow{EF}+\overrightarrow{FA} = 0$
Since $\overrightarrow{CD} = \overrightarrow{AF} = -\overrightarrow{FA}$,we have:
$\vec{a} + \vec{b} + \overrightarrow{CD} - \vec{a} + \overrightarrow{EF} - \overrightarrow{CD} = 0$
Alternatively,using the property of a regular hexagon,$\overrightarrow{FA} = \overrightarrow{CD} - \overrightarrow{BC} = \vec{a} - \vec{b}$.

(B) 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 proportional.
$\vec{AB} = (6-\alpha)\hat{i} + (11-10)\hat{j} + (11-13)\hat{k} = (6-\alpha)\hat{i} + 1\hat{j} - 2\hat{k}$.
$\vec{BC} = (\frac{9}{2}-6)\hat{i} + (\beta-11)\hat{j} + (-8-11)\hat{k} = -\frac{3}{2}\hat{i} + (\beta-11)\hat{j} - 19\hat{k}$.
For collinearity,$\frac{6-\alpha}{-3/2} = \frac{1}{\beta-11} = \frac{-2}{-19}$.
From $\frac{1}{\beta-11} = \frac{2}{19}$,we get $2(\beta-11) = 19 \Rightarrow 2\beta - 22 = 19 \Rightarrow 2\beta = 41 \Rightarrow 6\beta = 123$.
From $\frac{6-\alpha}{-3/2} = \frac{2}{19}$,we get $19(6-\alpha) = -3 \Rightarrow 114 - 19\alpha = -3 \Rightarrow 19\alpha = 117$.
Thus,$(19\alpha - 6\beta)^2 = (117 - 123)^2 = (-6)^2 = 36$.

(D) Given: $\vec{a}+\vec{b}+\vec{c}=\alpha \vec{d}$ ....$(i)$
Given: $\vec{b}+\vec{c}+\vec{d}=\beta \vec{a}$ ....(ii)
From $(i)$,$\vec{b}+\vec{c}=\alpha \vec{d}-\vec{a}$.
Substituting this into (ii): $(\alpha \vec{d}-\vec{a})+\vec{d}=\beta \vec{a}$.
Rearranging the terms: $(\alpha+1)\vec{d} = (\beta+1)\vec{a}$.
Since $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar,$\vec{d}$ must be a linear combination of $\vec{a}, \vec{b}, \vec{c}$. For the equation $(\alpha+1)\vec{d} = (\beta+1)\vec{a}$ to hold for non-coplanar vectors,the coefficients must be zero,implying $\alpha+1=0$ and $\beta+1=0$,so $\alpha=-1$ and $\beta=-1$.
Substituting $\alpha=-1$ into $(i)$: $\vec{a}+\vec{b}+\vec{c} = -\vec{d}$.
Therefore,$\vec{a}+\vec{b}+\vec{c}+\vec{d} = \vec{0}$.
Thus,$|\vec{a}+\vec{b}+\vec{c}+\vec{d}| = |\vec{0}| = 0$.

(A) Let $\hat{a} = \frac{\vec{a}}{|\vec{a}|}$ and $\hat{b} = \frac{\vec{b}}{|\vec{b}|}$ be the unit vectors along $\vec{a}$ and $\vec{b}$ respectively.
Then,$|\vec{b}| \vec{a} + |\vec{a}| \vec{b} = |\vec{a}| |\vec{b}| \left( \frac{\vec{a}}{|\vec{a}|} + \frac{\vec{b}}{|\vec{b}|} \right) = |\vec{a}| |\vec{b}| (\hat{a} + \hat{b})$.
Since $\hat{a} + \hat{b}$ is the diagonal of the rhombus formed by unit vectors $\hat{a}$ and $\hat{b}$,it represents the angle bisector of the angle between $\vec{a}$ and $\vec{b}$.
Thus,$|\vec{b}| \vec{a} + |\vec{a}| \vec{b}$ is a vector parallel to the angle bisector of $\vec{a}$ and $\vec{b}$.

(B) We are given $|\vec{f}|=10$,$|\vec{g}|=14$,and $|\vec{f}-\vec{g}|=15$.
Using the property $|\vec{f}-\vec{g}|^2 = |\vec{f}|^2 + |\vec{g}|^2 - 2(\vec{f} \cdot \vec{g})$,we have:
$15^2 = 10^2 + 14^2 - 2(\vec{f} \cdot \vec{g})$
$225 = 100 + 196 - 2(\vec{f} \cdot \vec{g})$
$225 = 296 - 2(\vec{f} \cdot \vec{g})$
$2(\vec{f} \cdot \vec{g}) = 296 - 225 = 71$.
Now,we need to find $|\vec{f}+\vec{g}|$.
Using the property $|\vec{f}+\vec{g}|^2 = |\vec{f}|^2 + |\vec{g}|^2 + 2(\vec{f} \cdot \vec{g})$,we substitute the known values:
$|\vec{f}+\vec{g}|^2 = 10^2 + 14^2 + 71$
$|\vec{f}+\vec{g}|^2 = 100 + 196 + 71 = 367$
Therefore,$|\vec{f}+\vec{g}| = \sqrt{367}$.

(D) Given that $|\vec{a}|=|\vec{b}|=|\vec{c}|=\sqrt{3}$.
Expanding the given equation $(\vec{a}+\vec{b}+\vec{c})^2+(\vec{b}+\vec{c}-\vec{a})^2+(\vec{c}+\vec{a}-\vec{b})^2=36$:
$3(|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2) + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 36$.
Substituting $|\vec{a}|^2=|\vec{b}|^2=|\vec{c}|^2=3$:
$3(3+3+3) + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 36$.
$27 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 36 \Rightarrow 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 9$.
Now,consider $|\vec{a}+\vec{b}+\vec{c}|^2 = |\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 3+3+3+9 = 18$.
We need to find $|2 \vec{a}-3 \vec{b}+2 \vec{c}|^2 = |2(\vec{a}+\vec{c})-3 \vec{b}|^2$.
From the expansion,we have $2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 9$.
Using the identity $|2 \vec{a}-3 \vec{b}+2 \vec{c}|^2 = 4|\vec{a}|^2 + 9|\vec{b}|^2 + 4|\vec{c}|^2 - 12(\vec{a} \cdot \vec{b}) - 12(\vec{b} \cdot \vec{c}) + 8(\vec{a} \cdot \vec{c})$.
Given the symmetry,$|2 \vec{a}-3 \vec{b}+2 \vec{c}|^2 = 4(3) + 9(3) + 4(3) - 12(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c}) + 8(\vec{a} \cdot \vec{c}) = 75$.

(D) Let the position vectors be $\vec{a} = \hat{i}-2 \hat{j}+3 \hat{k}$,$\vec{b} = 2 \hat{i}+3 \hat{j}+\hat{k}$,and $\vec{c} = -3 \hat{i}-\hat{j}-2 \hat{k}$.
Calculating the vectors for the sides:
$\overrightarrow{AB} = \vec{b} - \vec{a} = (2-1)\hat{i} + (3-(-2))\hat{j} + (1-3)\hat{k} = \hat{i} + 5\hat{j} - 2\hat{k}$.
$\overrightarrow{BC} = \vec{c} - \vec{b} = (-3-2)\hat{i} + (-1-3)\hat{j} + (-2-1)\hat{k} = -5\hat{i} - 4\hat{j} - 3\hat{k}$.
$\overrightarrow{AC} = \vec{c} - \vec{a} = (-3-1)\hat{i} + (-1-(-2))\hat{j} + (-2-3)\hat{k} = -4\hat{i} + \hat{j} - 5\hat{k}$.
Calculating the magnitudes:
$|\overrightarrow{AB}| = \sqrt{1^2 + 5^2 + (-2)^2} = \sqrt{1 + 25 + 4} = \sqrt{30}$.
$|\overrightarrow{BC}| = \sqrt{(-5)^2 + (-4)^2 + (-3)^2} = \sqrt{25 + 16 + 9} = \sqrt{50} = 5\sqrt{2}$.
$|\overrightarrow{AC}| = \sqrt{(-4)^2 + 1^2 + (-5)^2} = \sqrt{16 + 1 + 25} = \sqrt{42}$.
Since all side lengths are different,the points form a scalene triangle.

(D) Let the position vectors of vertices $P, Q, R$ be $\vec{p} = 4 \hat{i}+3 \hat{j}+6 \hat{k}$,$\vec{q} = 2 \hat{i}+2 \hat{j}+3 \hat{k}$,and $\vec{r} = 3 \hat{i}+\hat{j}+3 \hat{k}$.
First,calculate the lengths of sides $PQ$ and $PR$:
$PQ = |\vec{q} - \vec{p}| = |(2-4) \hat{i} + (2-3) \hat{j} + (3-6) \hat{k}| = |-2 \hat{i} - \hat{j} - 3 \hat{k}| = \sqrt{(-2)^2 + (-1)^2 + (-3)^2} = \sqrt{4+1+9} = \sqrt{14}$.
$PR = |\vec{r} - \vec{p}| = |(3-4) \hat{i} + (1-3) \hat{j} + (3-6) \hat{k}| = |-\hat{i} - 2 \hat{j} - 3 \hat{k}| = \sqrt{(-1)^2 + (-2)^2 + (-3)^2} = \sqrt{1+4+9} = \sqrt{14}$.
Since $PQ = PR$,$\triangle PQR$ is an isosceles triangle with $PQ = PR$.
In an isosceles triangle,the angle bisector of the vertex angle $(P)$ is also the median to the base $(QR)$.
Therefore,the point of intersection $A$ of the angle bisector of $P$ with $QR$ is the midpoint of $QR$.
$A = \frac{\vec{q} + \vec{r}}{2} = \frac{(2 \hat{i}+2 \hat{j}+3 \hat{k}) + (3 \hat{i}+\hat{j}+3 \hat{k})}{2} = \frac{5 \hat{i} + 3 \hat{j} + 6 \hat{k}}{2} = \frac{5}{2} \hat{i} + \frac{3}{2} \hat{j} + 3 \hat{k}$.

(A) Let $\vec{r}$ be the position vector of the point $C$.
Since point $C$ divides the line segment joining the points with position vectors $\vec{a} = 2 \hat{i}-3 \hat{j}+2 \hat{k}$ and $\vec{b} = 3 \hat{i}-\hat{j}-2 \hat{k}$ in the ratio $m:n = 2:3$,the position vector $\vec{r}$ is given by the section formula:
$\vec{r} = \frac{m\vec{b} + n\vec{a}}{m+n} = \frac{2(3 \hat{i}-\hat{j}-2 \hat{k}) + 3(2 \hat{i}-3 \hat{j}+2 \hat{k})}{2+3}$
$= \frac{(6 \hat{i}-2 \hat{j}-4 \hat{k}) + (6 \hat{i}-9 \hat{j}+6 \hat{k})}{5} = \frac{12 \hat{i}-11 \hat{j}+2 \hat{k}}{5} = \frac{12}{5} \hat{i}-\frac{11}{5} \hat{j}+\frac{2}{5} \hat{k}$.
Now,the distance of $C$ from the point $P$ with position vector $\vec{p} = 2 \hat{i}-\hat{j}+\hat{k}$ is given by $|\vec{r} - \vec{p}|$.
$\vec{r} - \vec{p} = (\frac{12}{5}-2) \hat{i} + (-\frac{11}{5}+1) \hat{j} + (\frac{2}{5}-1) \hat{k} = \frac{2}{5} \hat{i} - \frac{6}{5} \hat{j} - \frac{3}{5} \hat{k}$.
The distance $D = |\vec{r} - \vec{p}| = \sqrt{(\frac{2}{5})^2 + (-\frac{6}{5})^2 + (-\frac{3}{5})^2} = \sqrt{\frac{4}{25} + \frac{36}{25} + \frac{9}{25}} = \sqrt{\frac{49}{25}} = \frac{7}{5}$.

(D) Given: $\vec{a}=(2x+y)\hat{i}+3\hat{j}+9\hat{k}$ and $\vec{b}=2\hat{i}+\hat{j}-(x-y)\hat{k}$.
Since $\vec{a}$ and $\vec{b}$ are collinear,their components are proportional:
$\frac{2x+y}{2} = \frac{3}{1} = \frac{9}{-(x-y)}$.
From $\frac{2x+y}{2} = 3$,we get $2x+y=6$ $(i)$.
From $\frac{3}{1} = \frac{9}{y-x}$,we get $y-x=3$ or $y=x+3$ $(ii)$.
Substituting $(ii)$ into $(i)$:
$2x+(x+3)=6 \Rightarrow 3x=3 \Rightarrow x=1$.
Then $y=1+3=4$.
Finally,$x^3+27y^3 = (1)^3 + 27(4)^3 = 1 + 27(64) = 1 + 1728 = 1729$.

(C) Given $\vec{a}=2 \hat{i}+\hat{j}-\hat{k}$ and $\vec{b}=\hat{i}+3 \hat{j}-5 \hat{k}$.
First,calculate the vector $\vec{v} = 3 \vec{a}-2 \vec{b}$:
$\vec{v} = 3(2 \hat{i}+\hat{j}-\hat{k})-2(\hat{i}+3 \hat{j}-5 \hat{k}) = (6-2) \hat{i} + (3-6) \hat{j} + (-3+10) \hat{k} = 4 \hat{i}-3 \hat{j}+7 \hat{k}$.
Since $\vec{r}$ is along the vector $\vec{v}$,we can write $\vec{r} = x \vec{v} = x(4 \hat{i}-3 \hat{j}+7 \hat{k})$ for some scalar $x$.
Given $|\vec{r}| = \sqrt{74}$,we have $|x| \sqrt{4^2+(-3)^2+7^2} = \sqrt{74}$.
$|x| \sqrt{16+9+49} = \sqrt{74} \Rightarrow |x| \sqrt{74} = \sqrt{74} \Rightarrow |x| = 1$.
Since the direction of $\vec{r}$ is opposite to that of $\vec{v}$,we must have $x = -1$.
Therefore,$\vec{r} = -1(4 \hat{i}-3 \hat{j}+7 \hat{k}) = -4 \hat{i}+3 \hat{j}-7 \hat{k}$.

(B) Let the position vector of vertex $C$ be $\vec{c} = x\hat{i} + y\hat{j} + z\hat{k}$.
Given,centroid $G = \hat{i} = 1\hat{i} + 0\hat{j} + 0\hat{k}$.
We know that $G = \frac{\vec{A} + \vec{B} + \vec{C}}{3}$,where $\vec{A} = 2\hat{i} + \hat{j} + \hat{k}$ and $\vec{B} = 2\hat{i} + 4\hat{j} - 4\hat{k}$.
So,$3(1\hat{i} + 0\hat{j} + 0\hat{k}) = (2\hat{i} + \hat{j} + \hat{k}) + (2\hat{i} + 4\hat{j} - 4\hat{k}) + (x\hat{i} + y\hat{j} + z\hat{k})$.
$3\hat{i} = (4+x)\hat{i} + (5+y)\hat{j} + (-3+z)\hat{k}$.
Comparing coefficients: $4+x = 3 \Rightarrow x = -1$; $5+y = 0 \Rightarrow y = -5$; $-3+z = 0 \Rightarrow z = 3$.
Thus,$\vec{C} = -\hat{i} - 5\hat{j} + 3\hat{k}$.
Now,$AG^2 = |\vec{G} - \vec{A}|^2 = |(1-2)\hat{i} + (0-1)\hat{j} + (0-1)\hat{k}|^2 = |-1\hat{i} - 1\hat{j} - 1\hat{k}|^2 = 1+1+1 = 3$.
$BG^2 = |\vec{G} - \vec{B}|^2 = |(1-2)\hat{i} + (0-4)\hat{j} + (0+4)\hat{k}|^2 = |-1\hat{i} - 4\hat{j} + 4\hat{k}|^2 = 1+16+16 = 33$.
$CG^2 = |\vec{G} - \vec{C}|^2 = |(1+1)\hat{i} + (0+5)\hat{j} + (0-3)\hat{k}|^2 = |2\hat{i} + 5\hat{j} - 3\hat{k}|^2 = 4+25+9 = 38$.
Therefore,$AG^2 + BG^2 + CG^2 = 3 + 33 + 38 = 74$.

(C) Given,$\vec{a}=-2 \hat{i}+9 \hat{j}-6 \hat{k}$ and $\vec{b}=t \hat{i}-2 \hat{j}+6 \hat{k}$.
First,calculate the sum of the vectors: $\vec{a}+\vec{b} = (-2+t) \hat{i} + (9-2) \hat{j} + (-6+6) \hat{k} = (t-2) \hat{i} + 7 \hat{j} + 0 \hat{k}$.
Given that $|\vec{a}+\vec{b}|=25$,we have $\sqrt{(t-2)^2 + 7^2 + 0^2} = 25$.
Squaring both sides,we get $(t-2)^2 + 49 = 625$.
$(t-2)^2 = 625 - 49 = 576$.
Taking the square root,$t-2 = \pm 24$.
Case $1$: $t-2 = 24 \Rightarrow t = 26$.
Case $2$: $t-2 = -24 \Rightarrow t = -22$.
The sum of the values of $t$ is $26 + (-22) = 4$.

(D) Given position vectors are $\overrightarrow{OA} = 2\hat{i} + \hat{j} - \hat{k}$,$\overrightarrow{OB} = \hat{i} - 3\hat{j} + 5\hat{k}$,and $\overrightarrow{OC} = -3\hat{i} + 4\hat{j} + 4\hat{k}$.
Calculate the vectors representing the sides:
$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (1-2)\hat{i} + (-3-1)\hat{j} + (5-(-1))\hat{k} = -\hat{i} - 4\hat{j} + 6\hat{k}$.
$|\overrightarrow{AB}| = \sqrt{(-1)^2 + (-4)^2 + 6^2} = \sqrt{1 + 16 + 36} = \sqrt{53}$.
$\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = (-3-1)\hat{i} + (4-(-3))\hat{j} + (4-5)\hat{k} = -4\hat{i} + 7\hat{j} - \hat{k}$.
$|\overrightarrow{BC}| = \sqrt{(-4)^2 + 7^2 + (-1)^2} = \sqrt{16 + 49 + 1} = \sqrt{66}$.
$\overrightarrow{CA} = \overrightarrow{OA} - \overrightarrow{OC} = (2-(-3))\hat{i} + (1-4)\hat{j} + (-1-4)\hat{k} = 5\hat{i} - 3\hat{j} - 5\hat{k}$.
$|\overrightarrow{CA}| = \sqrt{5^2 + (-3)^2 + (-5)^2} = \sqrt{25 + 9 + 25} = \sqrt{59}$.
Since $|\overrightarrow{AB}| \neq |\overrightarrow{BC}| \neq |\overrightarrow{CA}|$,all three sides have different lengths.
Therefore,$\triangle ABC$ is a scalene triangle.

(B) Given position vectors are $\vec{OA} = 2 \hat{i} + 3 \hat{j} - \hat{k}$ and $\vec{OB} = \hat{i} - \hat{j} + 2 \hat{k}$.
First,we find the vector $\overrightarrow{AB} = \vec{OB} - \vec{OA} = (1-2)\hat{i} + (-1-3)\hat{j} + (2-(-1))\hat{k} = -\hat{i} - 4 \hat{j} + 3 \hat{k}$.
Next,we find the vector $\overrightarrow{BA} = \vec{OA} - \vec{OB} = (2-1)\hat{i} + (3-(-1))\hat{j} + (-1-2)\hat{k} = \hat{i} + 4 \hat{j} - 3 \hat{k}$.
The magnitude of $\overrightarrow{BA}$ is $|\overrightarrow{BA}| = \sqrt{1^2 + 4^2 + (-3)^2} = \sqrt{1 + 16 + 9} = \sqrt{26}$.
The unit vector along $\overrightarrow{BA}$ in the direction of $\overrightarrow{AB}$ is defined as $\frac{\overrightarrow{AB}}{|\overrightarrow{BA}|}$.
Therefore,the required unit vector is $\frac{-\hat{i} - 4 \hat{j} + 3 \hat{k}}{\sqrt{26}}$.

(A) Given position vectors are $\vec{a} = -3 \hat{i}-3 \hat{j}+4 \hat{k}$ and $\vec{b} = 4 \hat{i}-4 \hat{j}-3 \hat{k}$.
Since $L$ divides $OA$ in the ratio $2:3$,the position vector of $L$ is $\vec{OL} = \frac{2}{5} \vec{a} = \frac{2}{5}(-3 \hat{i}-3 \hat{j}+4 \hat{k})$.
Since $M$ divides $OB$ in the ratio $3:2$,the position vector of $M$ is $\vec{OM} = \frac{3}{5} \vec{b} = \frac{3}{5}(4 \hat{i}-4 \hat{j}-3 \hat{k})$.
Since $LP \parallel OB$ and $PM \parallel OA$,$OMPL$ is a parallelogram.
Therefore,the position vector of $P$ is $\vec{OP} = \vec{OL} + \vec{OM}$.
$\vec{OP} = \frac{2}{5}(-3 \hat{i}-3 \hat{j}+4 \hat{k}) + \frac{3}{5}(4 \hat{i}-4 \hat{j}-3 \hat{k})$
$\vec{OP} = \frac{1}{5}(-6 \hat{i}-6 \hat{j}+8 \hat{k} + 12 \hat{i}-12 \hat{j}-9 \hat{k})$
$\vec{OP} = \frac{1}{5}(6 \hat{i}-18 \hat{j}-\hat{k})$.
The distance from $O$ to $P$ is $|\vec{OP}| = \frac{1}{5} \sqrt{6^2 + (-18)^2 + (-1)^2} = \frac{1}{5} \sqrt{36 + 324 + 1} = \frac{\sqrt{361}}{5} = \frac{19}{5}$.

(D) Given vectors are $\vec{a}=3 \hat{i}+\hat{j}-2 \hat{k}$,$\vec{b}=-5 \hat{i}+7 \hat{j}$,and $\vec{c}=3 \hat{i}+y \hat{j}$.
First,calculate $\vec{a}-\vec{b}+\vec{c}$:
$\vec{a}-\vec{b}+\vec{c} = (3 \hat{i}+\hat{j}-2 \hat{k}) - (-5 \hat{i}+7 \hat{j}) + (3 \hat{i}+y \hat{j})$
$= (3+5+3) \hat{i} + (1-7+y) \hat{j} - 2 \hat{k}$
$= 11 \hat{i} + (y-6) \hat{j} - 2 \hat{k}$.
Now,find the magnitude $|\vec{a}-\vec{b}+\vec{c}| = \sqrt{11^2 + (y-6)^2 + (-2)^2} = \sqrt{121 + (y-6)^2 + 4} = \sqrt{125 + (y-6)^2}$.
Given that $|\vec{a}-\vec{b}+\vec{c}| = \sqrt{141}$,we have:
$\sqrt{125 + (y-6)^2} = \sqrt{141}$
$125 + (y-6)^2 = 141$
$(y-6)^2 = 141 - 125 = 16$
$y-6 = \pm 4$.
Thus,$y_1 = 6+4 = 10$ and $y_2 = 6-4 = 2$.
The value of $|y_1-y_2| = |10-2| = 8$.

(B) Given $\vec{a}=\hat{i}+2 \hat{j}-3 \hat{k}$ and $\vec{b}=2 \hat{i}-3 \hat{j}-5 \hat{k}$.
First,calculate $\vec{a}-\vec{b} = (1-2)\hat{i} + (2-(-3))\hat{j} + (-3-(-5))\hat{k} = -\hat{i} + 5\hat{j} + 2\hat{k}$.
The magnitude $|\vec{a}-\vec{b}| = \sqrt{(-1)^2 + 5^2 + 2^2} = \sqrt{1+25+4} = \sqrt{30}$.
The magnitude $|\vec{a}| = \sqrt{1^2 + 2^2 + (-3)^2} = \sqrt{1+4+9} = \sqrt{14}$.
The magnitude $|\vec{b}| = \sqrt{2^2 + (-3)^2 + (-5)^2} = \sqrt{4+9+25} = \sqrt{38}$.
Now,check the options. For option $B$,$|\vec{b}|-|\vec{a}| = \sqrt{38} - \sqrt{14} \approx 6.16 - 3.74 = 2.42$.
Since $|\vec{a}-\vec{b}| = \sqrt{30} \approx 5.47$,we have $5.47 > 2.42$.
Thus,$|\vec{a}-\vec{b}| > |\vec{b}|-|\vec{a}|$ is correct.

(A) Given position vectors are $\overrightarrow{OA} = \hat{i}-\hat{j}+2\hat{k}$ and $\overrightarrow{OB} = \hat{i}+2\hat{j}-2\hat{k}$.
Any point $C$ on the line passing through $A$ and $B$ can be represented as $\overrightarrow{OC} = \overrightarrow{OA} + t(\overrightarrow{OB} - \overrightarrow{OA})$.
$\overrightarrow{OB} - \overrightarrow{OA} = (\hat{i}+2\hat{j}-2\hat{k}) - (\hat{i}-\hat{j}+2\hat{k}) = 3\hat{j} - 4\hat{k}$.
So,$\overrightarrow{OC} = (\hat{i}-\hat{j}+2\hat{k}) + t(3\hat{j}-4\hat{k}) = \hat{i} + (3t-1)\hat{j} + (2-4t)\hat{k}$.
Given $BC = 10$,we have $|\overrightarrow{OC} - \overrightarrow{OB}| = 10$.
$\overrightarrow{OC} - \overrightarrow{OB} = (\hat{i} + (3t-1)\hat{j} + (2-4t)\hat{k}) - (\hat{i}+2\hat{j}-2\hat{k}) = (3t-3)\hat{j} + (4-4t)\hat{k}$.
$|\overrightarrow{OC} - \overrightarrow{OB}|^2 = (3t-3)^2 + (4-4t)^2 = 10^2 = 100$.
$9(t-1)^2 + 16(1-t)^2 = 100 \Rightarrow 25(t-1)^2 = 100 \Rightarrow (t-1)^2 = 4$.
$t-1 = \pm 2$,so $t = 3$ or $t = -1$.
For $t=3$,$\overrightarrow{OC} = \hat{i} + (3(3)-1)\hat{j} + (2-4(3))\hat{k} = \hat{i} + 8\hat{j} - 10\hat{k}$.

(D) Given $\vec{a} = t \vec{b}$ where $t < 0$ is a scalar.
Since $t$ is negative,the vectors $\vec{a}$ and $\vec{b}$ must point in opposite directions,which means they are unlike vectors.
Taking the modulus on both sides: $|\vec{a}| = |t \vec{b}| = |t| |\vec{b}|$.
Since $t < 0$,$|t|$ can be any positive real number.
If $|t| > 1$,then $|\vec{a}| > |\vec{b}|$.
If $|t| = 1$,then $|\vec{a}| = |\vec{b}|$.
If $0 < |t| < 1$,then $|\vec{a}| < |\vec{b}|$.
Therefore,$\vec{a}$ and $\vec{b}$ are unlike vectors,and the relationship between their magnitudes depends on the value of $|t|$,which can be either $|\vec{a}| \geq |\vec{b}|$ or $|\vec{a}| < |\vec{b}|$.

(C) The vector along the angle bisector of two vectors $\vec{a}$ and $\vec{b}$ is given by $\vec{c} = \lambda \left( \frac{\vec{a}}{|\vec{a}|} + \frac{\vec{b}}{|\vec{b}|} \right)$.
Given $\vec{a} = 2 \hat{i} - \hat{j} + 5 \hat{k}$,we have $|\vec{a}| = \sqrt{2^2 + (-1)^2 + 5^2} = \sqrt{4 + 1 + 25} = \sqrt{30}$.
Given $\vec{b} = 3 \hat{i} + 7 \hat{j} - \hat{k}$,we have $|\vec{b}| = \sqrt{3^2 + 7^2 + (-1)^2} = \sqrt{9 + 49 + 1} = \sqrt{59}$.
Let $x = \frac{1}{|\vec{a}|} = \frac{1}{\sqrt{30}}$ and $y = \frac{1}{|\vec{b}|} = \frac{1}{\sqrt{59}}$.
Then $\vec{c} = \lambda (x \vec{a} + y \vec{b}) = \lambda (x(2 \hat{i} - \hat{j} + 5 \hat{k}) + y(3 \hat{i} + 7 \hat{j} - \hat{k}))$.
Grouping the components,we get $\vec{c} = \lambda ((2x + 3y) \hat{i} + (7y - x) \hat{j} + (5x - y) \hat{k})$.
Comparing this with the options,option $C$ represents the vector along the bisector.

(A) Let the position vectors of vertices $A, B, C$ be $\vec{a}, \vec{b}, \vec{c}$ respectively. Using the section formula,the position vectors of $D, E, F$ are:
$\vec{d} = \frac{3\vec{b} + 2\vec{c}}{5}$,$\vec{e} = \frac{2\vec{c} + 1\vec{a}}{3}$,$\vec{f} = \frac{1\vec{a} + 3\vec{b}}{4}$.
Any point $P$ on $AD$ can be written as $\vec{p} = (1-t)\vec{a} + t\vec{d} = (1-t)\vec{a} + t(\frac{3\vec{b} + 2\vec{c}}{5}) = (1-t)\vec{a} + \frac{3t}{5}\vec{b} + \frac{2t}{5}\vec{c}$.
Since $P$ also lies on $BE$,$\vec{p} = (1-m)\vec{b} + m\vec{e} = (1-m)\vec{b} + m(\frac{2\vec{c} + \vec{a}}{3}) = \frac{m}{3}\vec{a} + (1-m)\vec{b} + \frac{2m}{3}\vec{c}$.
Comparing the coefficients of $\vec{a}, \vec{b}, \vec{c}$ (since $\vec{a}, \vec{b}, \vec{c}$ are linearly independent in the plane):
$1-t = \frac{m}{3}$,$\frac{3t}{5} = 1-m$,$\frac{2t}{5} = \frac{2m}{3}$.
From $\frac{2t}{5} = \frac{2m}{3}$,we get $3t = 5m$,so $m = \frac{3t}{5}$.
Substituting into $1-t = \frac{m}{3}$,we get $1-t = \frac{3t}{15} = \frac{t}{5}$,so $1 = \frac{6t}{5}$,which gives $t = \frac{5}{6}$.
Then $m = \frac{3}{5} \times \frac{5}{6} = \frac{1}{2}$.
Substituting $t = \frac{5}{6}$ into the expression for $\vec{p}$:
$\vec{p} = (1-\frac{5}{6})\vec{a} + \frac{3}{5}(\frac{5}{6})\vec{b} + \frac{2}{5}(\frac{5}{6})\vec{c} = \frac{1}{6}\vec{a} + \frac{1}{2}\vec{b} + \frac{1}{3}\vec{c}$.
Now,$\overrightarrow{AP} = \vec{p} - \vec{a} = \frac{1}{6}\vec{a} + \frac{1}{2}\vec{b} + \frac{1}{3}\vec{c} - \vec{a} = \frac{1}{2}\vec{b} + \frac{1}{3}\vec{c} - \frac{5}{6}\vec{a}$.
Since $\vec{a} + \vec{b} + \vec{c}$ is not the correct approach here,we use $\vec{a} + \vec{b} + \vec{c}$ is not needed,rather $\overrightarrow{AP} = \frac{1}{2}(\vec{b}-\vec{a}) + \frac{1}{3}(\vec{c}-\vec{a}) = \frac{1}{2}\overrightarrow{AB} + \frac{1}{3}\overrightarrow{AC}$.
Thus,$x_1 = \frac{1}{2}$ and $y_1 = \frac{1}{3}$.
Therefore,$x_1 + y_1 = \frac{1}{2} + \frac{1}{3} = \frac{5}{6}$.

(C) Given that $P$ and $Q$ lie on the curve $y = 2^{x+2}$.
For point $P$,$\overline{OP} \cdot \hat{i} = -1$,which implies the $x$-coordinate of $P$ is $x_P = -1$.
Substituting $x_P = -1$ into the curve equation: $y_P = 2^{-1+2} = 2^1 = 2$.
Thus,$P = (-1, 2)$ and $\overline{OP} = -\hat{i} + 2\hat{j}$.
For point $Q$,$\overline{OQ} \cdot \hat{i} = 2$,which implies the $x$-coordinate of $Q$ is $x_Q = 2$.
Substituting $x_Q = 2$ into the curve equation: $y_Q = 2^{2+2} = 2^4 = 16$.
Thus,$Q = (2, 16)$ and $\overline{OQ} = 2\hat{i} + 16\hat{j}$.
Now,calculate $\overline{OQ} - 4\overline{OP}$:
$\overline{OQ} - 4\overline{OP} = (2\hat{i} + 16\hat{j}) - 4(-\hat{i} + 2\hat{j})$
$= 2\hat{i} + 16\hat{j} + 4\hat{i} - 8\hat{j}$
$= (2+4)\hat{i} + (16-8)\hat{j}$
$= 6\hat{i} + 8\hat{j}$.

(B) Given the equation: $a+3 b+4 c=x(a-2 b+3 c)+y(a+5 b-2 c)+z(6 a+14 b+4 c)$.
Expanding the right side,we get: $a+3 b+4 c=a(x+y+6 z)+b(-2 x+5 y+14 z)+c(3 x-2 y+4 z)$.
Since $a, b, c$ are non-coplanar vectors,their coefficients must be equal on both sides.
Comparing the coefficients,we obtain the system of linear equations:
$x+y+6 z=1$ ...$(i)$
$-2 x+5 y+14 z=3$ ...(ii)
$3 x-2 y+4 z=4$ ...(iii)
Solving these equations:
From $(i)$,$x = 1 - y - 6z$.
Substituting into (ii): $-2(1-y-6z) + 5y + 14z = 3 \implies -2 + 2y + 12z + 5y + 14z = 3 \implies 7y + 26z = 5$ ...(iv).
Substituting into (iii): $3(1-y-6z) - 2y + 4z = 4 \implies 3 - 3y - 18z - 2y + 4z = 4 \implies -5y - 14z = 1$ ...$(v)$.
Solving (iv) and $(v)$: Multiply (iv) by $5$ and $(v)$ by $7$: $35y + 130z = 25$ and $-35y - 98z = 7$.
Adding them: $32z = 32 \implies z = 1$.
Substituting $z=1$ into $(v)$: $-5y - 14(1) = 1 \implies -5y = 15 \implies y = -3$.
Substituting $y=-3, z=1$ into $(i)$: $x - 3 + 6(1) = 1 \implies x + 3 = 1 \implies x = -2$.
Thus,$x+y+z = -2 - 3 + 1 = -4$.

(D) Two vectors $\vec{a}$ and $\vec{b}$ are collinear if $\vec{a} = \lambda \vec{b}$ for some scalar $\lambda$.
Given $\vec{a} = p\hat{i} - 2\hat{j} + 5\hat{k}$ and $\vec{b} = 1\hat{i} + q\hat{j} - 3\hat{k}$.
Equating components:
$p = \lambda(1) \implies p = \lambda$
$-2 = \lambda(q) \implies -2 = \lambda q$
$5 = \lambda(-3) \implies \lambda = -\frac{5}{3}$
Substituting $\lambda = -\frac{5}{3}$ into the other equations:
$p = -\frac{5}{3}$
$-2 = (-\frac{5}{3})q \implies q = -2 \times (-\frac{3}{5}) = \frac{6}{5}$
Thus,$p = -\frac{5}{3}$ and $q = \frac{6}{5}$.

(B) Given that vectors $a$ and $b$ are collinear,we can write $b = \lambda a$ for some scalar $\lambda$.
Taking the magnitude on both sides,we have $|b| = |\lambda| |a|$.
First,calculate the magnitude of vector $a$:
$|a| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$.
Given $|b| = 21$,substitute the values into the equation:
$21 = |\lambda| \times 7$.
$|\lambda| = \frac{21}{7} = 3$.
Thus,$\lambda = \pm 3$.
Substituting $\lambda$ back into the expression for $b$:
$b = \pm 3(2\hat{i} + 3\hat{j} + 6\hat{k}) = \pm(6\hat{i} + 9\hat{j} + 18\hat{k})$.

(A) The condition $|a + b| = |a| + |b|$ holds if and only if vectors $a$ and $b$ are in the same direction,meaning they are collinear and have the same sense.
This implies that $a = k b$ for some scalar $k > 0$.
Given $a = \hat{i} + x \hat{j} + \hat{k}$ and $b = \hat{i} + \hat{j} + \hat{k}$,we compare the components:
$\frac{1}{1} = \frac{x}{1} = \frac{1}{1}$.
From this,we get $x = 1$.
Since $k = 1 > 0$,the condition is satisfied for $x = 1$.

(C) Given the equation $x a + y b + z c = 0$,we substitute the vectors:
$x(\hat{i} + 2 \hat{j} + 3 \hat{k}) + y(2 \hat{i} + 3 \hat{j} + \hat{k}) + z(8 \hat{i} + 13 \hat{j} + 9 \hat{k}) = 0$
Equating the coefficients of $\hat{i}, \hat{j}, \hat{k}$ to zero,we get:
$x + 2y + 8z = 0$ $(i)$
$2x + 3y + 13z = 0$ $(ii)$
$3x + y + 9z = 0$ $(iii)$
From $(i) \times 2 - (ii)$,we get: $(2x + 4y + 16z) - (2x + 3y + 13z) = 0 \Rightarrow y + 3z = 0 \Rightarrow y = -3z$.
Substituting $y = -3z$ into $(i)$:
$x + 2(-3z) + 8z = 0 \Rightarrow x - 6z + 8z = 0 \Rightarrow x + 2z = 0 \Rightarrow x = -2z$.
Now,calculate the value of $\frac{xy}{z^2}$:
$\frac{xy}{z^2} = \frac{(-2z)(-3z)}{z^2} = \frac{6z^2}{z^2} = 6$.

(A) Let the vectors $a$ and $b$ be represented by sides $\vec{OA}$ and $\vec{OD}$ of a parallelogram $OADC$. The diagonal $\vec{OC} = a+b$ bisects the angle between $a$ and $b$. Let the angle between $a$ and $b$ be $2\theta$. Then the angle between $a$ and $a+b$ is $\theta$.
In the parallelogram $OADC$,we have $OA = |a|$ and $OD = AC = |b|$.
Consider the triangle $\triangle ABC$ formed by dropping a perpendicular from $C$ to the line $OA$ extended to $B$.
In $\triangle ABC$,$AB = |b| \cos 2\theta$ and $BC = |b| \sin 2\theta$.
In the right-angled triangle $\triangle OBC$,$\tan \theta = \frac{BC}{OB} = \frac{|b| \sin 2\theta}{|a| + |b| \cos 2\theta}$.
Using the double angle formulas $\sin 2\theta = \frac{2 \tan \theta}{1 + \tan^2 \theta}$ and $\cos 2\theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}$,we get:
$\tan \theta = \frac{|b| \left( \frac{2 \tan \theta}{1 + \tan^2 \theta} \right)}{|a| + |b| \left( \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \right)}$
$\tan \theta = \frac{2|b| \tan \theta}{|a|(1 + \tan^2 \theta) + |b|(1 - \tan^2 \theta)}$
Since $a$ and $b$ are non-collinear,$\theta \neq 0$,so $\tan \theta \neq 0$. Dividing by $\tan \theta$:
$1 = \frac{2|b|}{|a| + |a| \tan^2 \theta + |b| - |b| \tan^2 \theta}$
$|a| + |a| \tan^2 \theta + |b| - |b| \tan^2 \theta = 2|b|$
$|a|(1 + \tan^2 \theta) - |b|(1 + \tan^2 \theta) = 0$
$(|a| - |b|)(1 + \tan^2 \theta) = 0$
Since $1 + \tan^2 \theta = \sec^2 \theta \neq 0$,we must have $|a| - |b| = 0$,which implies $|a| = |b|$.

(D) Given,$\vec{OA}=\vec{a}, \vec{OD}=\vec{d}$.
Since $OA \parallel CB$ and $\frac{OA}{CB}=2$,we have $\vec{CB} = \frac{1}{2}\vec{OA} = \frac{1}{2}\vec{a}$.
Since $OD \parallel AB$ and $\frac{OD}{AB}=\frac{1}{3}$,we have $\vec{AB} = 3\vec{OD} = 3\vec{d}$.
Now,we express the required sum in terms of vectors $\vec{a}$ and $\vec{d}$:
$\vec{AD} = \vec{OD} - \vec{OA} = \vec{d} - \vec{a}$.
$\vec{OC} = \vec{OD} + \vec{DC}$.
From the pentagon geometry,$\vec{OC} = \vec{OA} + \vec{AB} + \vec{BC} = \vec{a} + 3\vec{d} - \frac{1}{2}\vec{a} = \frac{1}{2}\vec{a} + 3\vec{d}$.
Also,$\vec{DC} = \vec{OC} - \vec{OD} = (\frac{1}{2}\vec{a} + 3\vec{d}) - \vec{d} = \frac{1}{2}\vec{a} + 2\vec{d}$.
Summing these: $\vec{AD} + \vec{OC} + \vec{DC} = (\vec{d} - \vec{a}) + (\frac{1}{2}\vec{a} + 3\vec{d}) + (\frac{1}{2}\vec{a} + 2\vec{d})$.
$= (-\vec{a} + \frac{1}{2}\vec{a} + \frac{1}{2}\vec{a}) + (\vec{d} + 3\vec{d} + 2\vec{d}) = 0\vec{a} + 6\vec{d} = 6\vec{d}$.

(A) Given that $D, E$,and $F$ are the mid-points of $AB, AC$,and $BC$ respectively in $\triangle ABC$.
Let the position vectors of vertices $A, B$,and $C$ be $\vec{a}, \vec{b}$,and $\vec{c}$ respectively.
Then,the position vectors of the mid-points are:
$\overrightarrow{D} = \frac{\vec{a} + \vec{b}}{2}$
$\overrightarrow{E} = \frac{\vec{a} + \vec{c}}{2}$
$\overrightarrow{F} = \frac{\vec{b} + \vec{c}}{2}$
Now,we calculate the vectors $\overrightarrow{BE}$ and $\overrightarrow{AF}$:
$\overrightarrow{BE} = \overrightarrow{E} - \overrightarrow{B} = \frac{\vec{a} + \vec{c}}{2} - \vec{b} = \frac{\vec{a} + \vec{c} - 2\vec{b}}{2}$
$\overrightarrow{AF} = \overrightarrow{F} - \overrightarrow{A} = \frac{\vec{b} + \vec{c}}{2} - \vec{a} = \frac{\vec{b} + \vec{c} - 2\vec{a}}{2}$
Adding these two vectors:
$\overrightarrow{BE} + \overrightarrow{AF} = \frac{\vec{a} + \vec{c} - 2\vec{b} + \vec{b} + \vec{c} - 2\vec{a}}{2}$
$= \frac{2\vec{c} - \vec{a} - \vec{b}}{2} = \vec{c} - \frac{\vec{a} + \vec{b}}{2}$
Since $\overrightarrow{DC} = \overrightarrow{C} - \overrightarrow{D} = \vec{c} - \frac{\vec{a} + \vec{b}}{2}$,we have:
$\overrightarrow{BE} + \overrightarrow{AF} = \overrightarrow{DC}$

(D) vector $\vec{r} = a\hat{i} + b\hat{j} + c\hat{k}$ is equally inclined to the coordinate axes if its direction cosines are equal,i.e.,$\cos \alpha = \cos \beta = \cos \gamma$.
This implies that the direction ratios must be equal in magnitude,i.e.,$|a| = |b| = |c|$.
Checking the options:
For option $D$,$\vec{v} = 4\hat{i} + 4\hat{j} + 4\hat{k}$.
The direction ratios are $a=4, b=4, c=4$.
The magnitude is $|\vec{v}| = \sqrt{4^2 + 4^2 + 4^2} = \sqrt{48} = 4\sqrt{3}$.
The direction cosines are $\left(\frac{4}{4\sqrt{3}}, \frac{4}{4\sqrt{3}}, \frac{4}{4\sqrt{3}}\right) = \left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$.
Since all direction cosines are equal,the vector $4\hat{i} + 4\hat{j} + 4\hat{k}$ is equally inclined with the coordinate axes.