The distance of the point ($$-1,9,-16$$) from the plane

$$2x+3y-z=5$$ measured parallel to the line

$${{x + 4} \over 3} = {{2 - y} \over 4} = {{z - 3} \over {12}}$$ is :

For three positive integers p, q, r, $${x^{p{q^2}}} = {y^{qr}} = {z^{{p^2}r}}$$ and r = pq + 1 such that 3, 3 log$$_yx$$, 3 log$$_zy$$, 7 log$$_xz$$ are in A.P. with common difference $$\frac{1}{2}$$. Then r-p-q is equal to

Let PQR be a triangle. The points A, B and C are on the sides QR, RP and PQ respectively such that

$${{QA} \over {AR}} = {{RB} \over {BP}} = {{PC} \over {CQ}} = {1 \over 2}$$. Then $${{Area(\Delta PQR)} \over {Area(\Delta ABC)}}$$ is equal to :

Let $$f(x) = \left\{ {\matrix{ {{x^2}\sin \left( {{1 \over x}} \right)} & {,\,x \ne 0} \cr 0 & {,\,x = 0} \cr } } \right.$$

Then at $$x=0$$