If the foot of the perpendicular from the point $$\mathrm{A}(-1,4,3)$$ on the plane $$\mathrm{P}: 2 x+\mathrm{m} y+\mathrm{n} z=4$$, is $$\left(-2, \frac{7}{2}, \frac{3}{2}\right)$$, then the distance of the point A from the plane P, measured parallel to a line with direction ratios $$3,-1,-4$$, is equal to :

Let $$\overrightarrow{\mathrm{a}}=3 \hat{i}+\hat{j}$$ and $$\overrightarrow{\mathrm{b}}=\hat{i}+2 \hat{j}+\hat{k}$$. Let $$\overrightarrow{\mathrm{c}}$$ be a vector satisfying $$\overrightarrow{\mathrm{a}} \times(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}})=\overrightarrow{\mathrm{b}}+\lambda \overrightarrow{\mathrm{c}}$$. If $$\overrightarrow{\mathrm{b}}$$ and $$\overrightarrow{\mathrm{c}}$$ are non-parallel, then the value of $$\lambda$$ is :

The angle of elevation of the top of a tower from a point A due north of it is $$\alpha$$ and from a point B at a distance of 9 units due west of A is $$\cos ^{-1}\left(\frac{3}{\sqrt{13}}\right)$$. If the distance of the point B from the tower is 15 units, then $$\cot \alpha$$ is equal to :

The statement $$(p \wedge q) \Rightarrow(p \wedge r)$$ is equivalent to :