If $D, E, F$ are the midpoints of the sides $BC, CA$ and $AB$ of the triangle $ABC$,then $\overrightarrow{AD} + \overrightarrow{BE} + \overrightarrow{CF}$ is:

If two vertices of a triangle are $\hat{i} - \hat{j}$ and $\hat{j} + \hat{k}$,then the third vertex can be

The vector $\frac{1}{3} (2\hat{i} - 2\hat{j} + \hat{k})$ is ....

If $\hat{i}+4 \hat{j}+3 \hat{k}$,$\hat{i}+2 \hat{j}+3 \hat{k}$,and $3 \hat{i}+2 \hat{j}+\hat{k}$ are position vectors of $A$,$B$,and $C$ respectively,and if $D$ and $E$ are midpoints of sides $BC$ and $AC$,then $\overrightarrow{DE}$ is equal to:

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

If $\vec{a}=-2 \hat{i}+9 \hat{j}-6 \hat{k}$ and $\vec{b}=t \hat{i}-2 \hat{j}+6 \hat{k}$ are vectors such that $|\vec{a}+\vec{b}|=25$,then the sum of the values of $t$ is