The distance of the point having position vector $$ - \widehat i + 2\widehat j + 6\widehat k$$ from the straight line passing through the point (2, 3, – 4) and parallel to the vector, $$6\widehat i + 3\widehat j - 4\widehat k$$ is :

Let A (3, 0, –1), B(2, 10, 6) and C(1, 2, 1) be the vertices of a triangle and M be the midpoint of AC. If G divides BM in the ratio, 2 : 1, then cos ($$\angle $$GOA) (O being the origin) is equal to :

If a unit vector $$\overrightarrow a $$ makes angles $$\pi $$/3 with $$\widehat i$$ , $$\pi $$/ 4 with $$\widehat j$$ and $$\theta $$$$ \in $$(0, $$\pi $$) with $$\widehat k$$, then a value of $$\theta $$ is :-

Let $$\overrightarrow \alpha = 3\widehat i + \widehat j$$ and $$\overrightarrow \beta = 2\widehat i - \widehat j + 3 \widehat k$$ . If $$\overrightarrow \beta = {\overrightarrow \beta _1} - \overrightarrow {{\beta _2}} $$, where $${\overrightarrow \beta _1}$$ is parallel to $$\overrightarrow \alpha $$ and $$\overrightarrow {{\beta _2}} $$ is perpendicular to $$\overrightarrow \alpha $$ , then $${\overrightarrow \beta _1} \times \overrightarrow {{\beta _2}} $$ is equal to