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

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

If the sum of two unit vectors is a unit vector, then the magnitude of their difference is

If $2 \hat{i}+4 \hat{j}-5 \hat{k}$, $\hat{i}+\hat{j}+\hat{k}$, and $\hat{j}+2 \hat{k}$ are the position vectors of the vertices $A$, $B$, and $C$ of a triangle respectively, then a unit vector along the median drawn through the vertex $A$ is

If $C$ is the midpoint of $AB$ and $P$ is any point outside $AB$,then

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 $5 \alpha-9 \beta+13 \gamma=$