If $PQ + QR = (2\lambda^2 - 5)RP$,then $\lambda$ is equal to

If the magnitudes of $\bar{a}$,$\bar{b}$ and $\bar{a}+\bar{b}$ are respectively $3$,$4$ and $5$,then the magnitude of $\bar{a}-\bar{b}$ is

$ABC$ is an isosceles triangle right-angled at $A$. Forces of magnitude $2\sqrt{2}$,$5$,and $6$ act along $\overrightarrow{BC}$,$\overrightarrow{CA}$,and $\overrightarrow{AB}$ respectively. The magnitude of their resultant force is

The points in space represented by the position vectors $A = 4\hat{i}+\hat{j}+3\hat{k}$,$B = 6\hat{i}-2\hat{j}-3\hat{k}$,and $C = \hat{i}-\hat{j}-3\hat{k}$ form:

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

The points with position vectors $60\,i + 3\,j$,$40\,i - 8\,j$,and $a\,i - 52\,j$ are collinear,if $a = $