Let $$\vec{a}$$ be a non-zero vector parallel to the line of intersection of the two planes described by $$\hat{i}+\hat{j}, \hat{i}+\hat{k}$$ and $$\hat{i}-\hat{j}, \hat{j}-\hat{k}$$. If $$\theta$$ is the angle between the vector $$\vec{a}$$ and the vector $$\vec{b}=2 \hat{i}-2 \hat{j}+\hat{k}$$ and $$\vec{a} \cdot \vec{b}=6$$, then the ordered pair $$(\theta,|\vec{a} \times \vec{b}|)$$ is equal to :

Let $$\vec{a}=2 \hat{i}+7 \hat{j}-\hat{k}, \vec{b}=3 \hat{i}+5 \hat{k}$$ and $$\vec{c}=\hat{i}-\hat{j}+2 \hat{k}$$. Let $$\vec{d}$$ be a vector which is perpendicular to both $$\vec{a}$$ and $$\vec{b}$$, and $$\vec{c} \cdot \vec{d}=12$$. Then $$(-\hat{i}+\hat{j}-\hat{k}) \cdot(\vec{c} \times \vec{d})$$ is equal to :

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 :