Let the position vectors of three vertices of a triangle be $4 \overrightarrow{p} + \overrightarrow{q} - 3 \overrightarrow{r}$,$-5 \overrightarrow{p} + \overrightarrow{q} + 2 \overrightarrow{r}$,and $2 \overrightarrow{p} - \overrightarrow{q} + 2 \overrightarrow{r}$. If the position vectors of the orthocenter $(O)$ and the circumcenter $(C)$ of the triangle are $\frac{\overrightarrow{p} + \overrightarrow{q} + \overrightarrow{r}}{4}$ and $\alpha \overrightarrow{p} + \beta \overrightarrow{q} + \gamma \overrightarrow{r}$ respectively,then $\alpha + 2 \beta + 5 \gamma$ is equal to:

If $a=2 \hat{i}+3 \hat{j}+\hat{k}$, $b=\hat{i}-3 \hat{j}-5 \hat{k}$ and $c=3 \hat{i}-4 \hat{k}$, then match the items of List-$I$ with those of List-$II$.

$A$. Unit vector in the direction opposite to that $a-b$ is	$(i) \ 5 \hat{i} + 3 \hat{j} - 3 \hat{k}$
$B$. If $\vec{AB} = a, \vec{BC} = b$, then $\vec{CA} =$	$(ii) \ 2 \hat{i} - \frac{8}{3} \hat{k}$
$C$. If $a, b, c$ are the position vectors of the vertices of a triangle then, its centroid is	$(iii) \ -3 \hat{i} + 4 \hat{k}$
$D$. If $d$ is a vector of magnitude $2 \sqrt{14}$ and parallel to the vector $a$, then $b + d =$	$(iv) \ -\frac{\hat{i}}{\sqrt{73}} - \frac{6 \hat{j}}{\sqrt{73}} - \frac{6 \hat{k}}{\sqrt{73}}$
	$(v) \ 3 \hat{i} + 5 \hat{j} - 3 \hat{k}$

Let $\vec{a} = 2\hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + \hat{k}$ be two vectors. Consider a vector $\vec{c} = \alpha\vec{a} + \beta\vec{b}$,where $\alpha, \beta \in \mathbb{R}$. If the projection of $\vec{c}$ on the vector $(\vec{a} + \vec{b})$ is $3\sqrt{2}$,then the minimum value of $(\vec{c} - (\vec{a} \times \vec{b})) \cdot \vec{c}$ is equal to:

Let $O$ be the origin and $\overline{OA} = 2\hat{i} + 2\hat{j} + \hat{k}$,$\overline{OB} = \hat{i} - 2\hat{j} + 2\hat{k}$ and $\overline{OC} = \frac{1}{2}(\overline{OB} - \lambda\overline{OA})$ for some $\lambda > 0$. If $|\overline{OB} \times \overline{OC}| = \frac{9}{2}$,then which of the following statements is (are) $TRUE$?
$(A)$ Projection of $\overline{OC}$ on $\overline{OA}$ is $-\frac{3}{2}$
$(B)$ Area of the triangle $OAB$ is $\frac{9}{2}$
$(C)$ Area of the triangle $ABC$ is $\frac{9}{2}$
$(D)$ The acute angle between the diagonals of the parallelogram with adjacent sides $\overline{OA}$ and $\overline{OC}$ is $\frac{\pi}{3}$

If $\vec{a} = \lambda x \hat{i} + y \hat{j} + 4z \hat{k}$,$\vec{b} = y \hat{i} + x \hat{j} + 3y \hat{k}$,and $\vec{c} = -z \hat{i} - 2z \hat{j} - (\lambda + 1) \hat{k}$ are the sides of the triangle $ABC$,where $x, y, z$ are not all zero,such that $\vec{a} + \vec{b} + \vec{c} = \vec{0}$,then the value of $\lambda$ is:

The point $B$ divides the arc $AC$ of a quadrant of a circle in the ratio $1 : 2$. If $O$ is the centre and $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$,then the vector $\overrightarrow{OC}$ is