The area (in sq. units) of the parallelogram whose diagonals are along the vectors $$8\widehat i - 6\widehat j$$ and $$3\widehat i + 4\widehat j - 12\widehat k,$$ is :

Let $$\overrightarrow a = 2\widehat i + \widehat j -2 \widehat k$$ and $$\overrightarrow b = \widehat i + \widehat j$$.

Let $$\overrightarrow c $$ be a vector such that $$\left| {\overrightarrow c - \overrightarrow a } \right| = 3$$,

$$\left| {\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c } \right| = 3$$ and the angle between $$\overrightarrow c $$ and $\overrightarrow a \times \overrightarrow b$ is $$30^\circ $$.

Then $$\overrightarrow a .\overrightarrow c $$ is equal to :

Let ABC be a triangle whose circumcentre is at P. If the position vectors of A, B, C and P are $$\overrightarrow a ,\overrightarrow b ,\overrightarrow c $$ and $${{\overrightarrow a + \overrightarrow b + \overrightarrow c } \over 4}$$ respectively, then the position vector of the orthocentre of this triangle, is :

In a triangle ABC, right angled at the vertex A, if the position vectors of A, B and C are respectively 3$$\widehat i$$ + $$\widehat j$$ $$-$$ $$\widehat k$$,   $$-$$$$\widehat i$$ + 3$$\widehat j$$ + p$$\widehat k$$ and 5$$\widehat i$$ + q$$\widehat j$$ $$-$$ 4$$\widehat k$$, then the point (p, q) lies on a line :