If one side of a square is represented by the vector $3i + 4j + 5k,$ then the area of the square is

In the given figure (a square),identify the following vectors:
Collinear but not equal

If $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$,$\vec{b} = 2\hat{i} + 3\hat{j} + \hat{k}$,$\vec{c} = 3\hat{i} + \hat{j} + 2\hat{k}$ and $\alpha \vec{a} + \beta \vec{b} + \gamma \vec{c} = -3(\hat{i} - \hat{k})$. Then the triplet $(\alpha, \beta, \gamma)$ is

If $D, E, F$ are respectively the midpoints of $AB, AC$ and $BC$ in $\Delta ABC$,then $\overrightarrow{BE} + \overrightarrow{AF} = $

Let $\overrightarrow{a}$ and $\overrightarrow{b}$ be two vectors such that $|\overrightarrow{a}| = 2$ and $|\overrightarrow{b}| = 3$. Then the ratio of the projection of $\overrightarrow{a}$ on $\overrightarrow{b}$ to that of $\overrightarrow{b}$ on $\overrightarrow{a}$ is:

If the points with position vectors $(\alpha \hat{i}+10 \hat{j}+13 \hat{k})$,$(6 \hat{i}+11 \hat{j}+11 \hat{k})$,and $(\frac{9}{2} \hat{i}+\beta \hat{j}-8 \hat{k})$ are collinear,then $(19 \alpha-6 \beta)^2=$