If $\vec{a}=4 \hat{i}+13 \hat{j}-18 \hat{k}$,$\vec{b}=\hat{i}-2 \hat{j}+3 \hat{k}$,and $\vec{c}=2 \hat{i}+3 \hat{j}-4 \hat{k}$ are three vectors such that $\vec{a}=x \vec{b}+y \vec{c}$,then $x+y=$

The perimeter of the triangle whose vertices have the position vectors $(i + j + k)$,$(5i + 3j - 3k)$,and $(2i + 5j + 9k)$ is given by:

$ABCD$ is a parallelogram such that $L$ is the mid-point of $BC$. Then,$\vec{AL}$ is equal to:

The unit vector in the direction of the vector $(1, 0, 0)$ is $.......$

Let a vector $\alpha \hat{i}+\beta \hat{j}$ be obtained by rotating the vector $\sqrt{3} \hat{i}+\hat{j}$ by an angle $45^{\circ}$ about the origin in the counterclockwise direction in the first quadrant. Then the area of the triangle having vertices $(\alpha, \beta), (0, \beta)$ and $(0,0)$ is equal to

Let $ABCD$ be a parallelogram and $2\hat{i}+\hat{j}$,$4\hat{i}+5\hat{j}+4\hat{k}$ and $-\hat{i}-4\hat{j}-3\hat{k}$ be the position vectors of the vertices $A$,$B$,and $D$ respectively. Then the position vector of one of the points of trisection of the diagonal $AC$ is