An arc PQ of a circle subtends a right angle at its centre O. The mid point of the arc PQ is R. If $$\overrightarrow {OP} = \overrightarrow u ,\overrightarrow {OR} = \overrightarrow v $$, and $$\overrightarrow {OQ} = \alpha \overrightarrow u + \beta \overrightarrow v $$, then $$\alpha ,{\beta ^2}$$ are the roots of the equation :

Let the vectors $$\vec{u}_{1}=\hat{i}+\hat{j}+a \hat{k}, \vec{u}_{2}=\hat{i}+b \hat{j}+\hat{k}$$ and $$\vec{u}_{3}=c \hat{i}+\hat{j}+\hat{k}$$ be coplanar. If the vectors $$\vec{v}_{1}=(a+b) \hat{i}+c \hat{j}+c \hat{k}, \vec{v}_{2}=a \hat{i}+(b+c) \hat{j}+a \hat{k}$$ and $$\vec{v}_{3}=b \hat{i}+b \hat{j}+(c+a) \hat{k}$$ are also coplanar, then $$6(\mathrm{a}+\mathrm{b}+\mathrm{c})$$ is equal to :

The area of the quadrilateral $$\mathrm{ABCD}$$ with vertices $$\mathrm{A}(2,1,1), \mathrm{B}(1,2,5), \mathrm{C}(-2,-3,5)$$ and $$\mathrm{D}(1,-6,-7)$$ is equal to :

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