The position vector of a point lying on the line joining the points whose position vectors are $\hat{i}+\hat{j}-\hat{k}$ and $\hat{i}-\hat{j}+\hat{k}$ is:

If $\alpha, \beta, \gamma$ are real numbers such that $(\frac{7}{3}+\beta) \hat{i}-\hat{j}+(\alpha+\gamma) \hat{k}=\frac{5}{3}(\alpha \hat{i}+\hat{j}-\hat{k})+\beta(2 \hat{j}+\hat{k})+(\hat{i}+\gamma \hat{j}+3 \hat{k})$,then $5 \alpha-9 \beta+13 \gamma=$

Let $u$ and $v$ be two vectors in $R^2$. If $|u+v|^2=2(|u|^2+|v|^2)$,then .....

If $\overrightarrow {AO} + \overrightarrow {OB} = \overrightarrow {BO} + \overrightarrow {OC} ,$ then $A, B, C$ form

Three non-zero non-collinear vectors $\vec{a}, \vec{b}$ and $\vec{c}$ are such that $\vec{a}+3\vec{b}$ is collinear with $\vec{c}$,and $3\vec{b}+2\vec{c}$ is collinear with $\vec{a}$. Then $\vec{a}+3\vec{b}+2\vec{c}$ equals to

If $a = 3i - 2j + k$,$b = 2i - 4j - 3k$ and $c = -i + 2j + 2k$,then $a + b + c$ is