If $A(1, 0, 0)$,$B(0, 1, 0)$,and $C(0, 0, 1)$ are given,and $\vec{AB} = \vec{CX}$,then the point $X$ is:

Let $A, B, C$ be three points whose position vectors are $\overrightarrow{a} = \hat{i} + 4\hat{j} + 3\hat{k}$,$\overrightarrow{b} = 2\hat{i} + \alpha\hat{j} + 4\hat{k}$ (where $\alpha \in R$),and $\overrightarrow{c} = 3\hat{i} - 2\hat{j} + 5\hat{k}$. If $\alpha$ is the smallest positive integer for which $\vec{a}, \vec{b}, \vec{c}$ are non-collinear,then the length of the median in $\triangle ABC$ through $A$ is:

$P$ is the point of intersection of the diagonals of the parallelogram $ABCD$. If $O$ is any point,then $\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} + \overrightarrow{OD} = $

If the points whose position vectors are $2 \hat{i}+\hat{j}+\hat{k}$, $6 \hat{i}-\hat{j}+2 \hat{k}$, and $14 \hat{i}-5 \hat{j}+p \hat{k}$ are collinear, then the value of $p$ is:

$D, E, F$ are points on the sides $BC, CA$ and $AB$ of a $\triangle ABC$ respectively,dividing them in the ratios $2:3, 1:2, 3:1$ internally. The lines $BE$ and $CF$ intersect on the line $AD$ at $P$. If $\overrightarrow{AP} = x_1 \overrightarrow{AB} + y_1 \overrightarrow{AC}$,then $x_1 + y_1 =$

If the vector $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ makes angles $\alpha, \beta, \gamma$ with vectors $\hat{i}, \hat{j}, \hat{k}$ respectively,then: