Classify the following measure as a scalar or a vector:
$40^{o}$

Using vectors,find the value of $k$ such that the points $(k,-10,3), (1,-1,3)$ and $(3,5,3)$ are collinear.

If the position vectors of the points $A, B, C, D$ given by $\hat{i}+2 \hat{j}+3 \hat{k}, 2 \hat{i}-\hat{j}+2 \hat{k}$, $\frac{1}{4}(7 \hat{i}+15 \hat{j}+15 \hat{k})$ and $\frac{1}{3}[7 \hat{i}+2 \hat{j}+(5+3 a) \hat{k}]$ respectively are such that $|AC|=|BD|$, then $16(3a-1)^2=$

Let $\overrightarrow{OA} = \hat{i} - 3\hat{j} + \hat{k}$, $\overrightarrow{OB} = \hat{i} + 3\hat{j} - 2\hat{k}$, and $\overrightarrow{OC} = 4\hat{i} + 3\hat{j} + 5\hat{k}$ be the position vectors of three points $A$, $B$, and $C$. Let $P$ be the point which divides $AB$ in the ratio $2:1$. If $l, m, n$ are the direction cosines of the vector $\overrightarrow{PC}$, then $l + 3m + 2n =$

If $ABCDEF$ is a regular hexagon,then $\overrightarrow {AD} + \overrightarrow {EB} + \overrightarrow {FC} = $

Let $\vec{a}$ and $\vec{b}$ be non-collinear vectors. If the vectors $(\lambda-1) \vec{a}+2 \vec{b}$ and $3 \vec{a}+\lambda \vec{b}$ are collinear, then the set of all possible values of $\lambda$ is