If the points $$\mathrm{P}$$ and $$\mathrm{Q}$$ are respectively the circumcenter and the orthocentre of a $$\triangle \mathrm{ABC}$$, then $$\overrightarrow{\mathrm{PA}}+\overrightarrow{\mathrm{PB}}+\overrightarrow{\mathrm{PC}}$$ is equal to :

Let O be the origin and the position vector of the point P be $$ - \widehat i - 2\widehat j + 3\widehat k$$. If the position vectors of the points A, B and C are $$ - 2\widehat i + \widehat j - 3\widehat k,2\widehat i + 4\widehat j - 2\widehat k$$ and $$ - 4\widehat i + 2\widehat j - \widehat k$$ respectively, then the projection of the vector $$\overrightarrow {OP} $$ on a vector perpendicular to the vectors $$\overrightarrow {AB} $$ and $$\overrightarrow {AC} $$ is :

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 :