If $$\mathrm{D}, \mathrm{E}$$ and $$\mathrm{F}$$ are the mid-points of the sides $$\mathrm{BC}$$, $$\mathrm{CA}$$ and $$\mathrm{AB}$$ of triangle $$\mathrm{ABC}$$ respectively, then $$\overline{\mathrm{AD}}+\frac{2}{3} \overline{\mathrm{BE}}+\frac{1}{3} \overline{\mathrm{CF}}=$$

If two vertices of a triangle are $$\mathrm{A}(3,1,4)$$ and $$\mathrm{B}(-4,5,-3)$$ and the centroid of the triangle is $$G(-1,2,1)$$, then the third vertex $$C$$ of the triangle is

Let two non-collinear vectors $$\hat{a}$$ and $$\hat{b}$$ form an acute angle. A point $$\mathrm{P}$$ moves, so that at any time $$t$$ the position vector $$\overline{\mathrm{OP}}$$, where $$\mathrm{O}$$ is origin, is given by $$\hat{a} \sin t+\hat{b} \cos t$$, when $$P$$ is farthest from origin $$O$$, let $$M$$ be the length of $$\overline{\mathrm{OP}}$$ and $$\hat{\mathrm{u}}$$ be the unit vector along $$\overline{\mathrm{OP}}$$, then

The distance of the point having position vector $$\hat{i}-2 \hat{j}-6 \hat{k}$$, from the straight line passing through the point $$(2,-3,-4)$$ and parallel to the vector $$6 \hat{i}+3 \hat{j}-4 \hat{k}$$ is units.