In a parallelogram $ABCD$,if the diagonals are $\overline{AC}$ and $\overline{BD}$,then $\overline{AC} - \overline{BD} = \dots$

If $P$ and $Q$ are two points on the curve $y = 2^{x+2}$ in the rectangular Cartesian coordinate system such that $\overline{OP} \cdot \hat{i} = -1$ and $\overline{OQ} \cdot \hat{i} = 2$,then $\overline{OQ} - 4\overline{OP} = $

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

Let $\vec{a}, \vec{b}, \vec{c}$ be three non-zero vectors such that no two of them are collinear. If the vector $\vec{a} + 3\vec{b}$ is collinear with $\vec{c}$ and $\vec{b} + 2\vec{c}$ is collinear with $\vec{a}$,then find the value of $\vec{a} + 3\vec{b} + 6\vec{c}$.

If the position vectors of the points $A, B, C$ are $a, b, 3a - 2b$ respectively,then the points $A, B, C$ are

$L$ and $M$ are two points with position vectors $2 \vec{a}-\vec{b}$ and $\vec{a}+2 \vec{b}$ respectively. The position vector of the point $N$ which divides the line segment $LM$ in the ratio $2:1$ externally is