Consider the set of eight vectors $$V = \left\{ {a\,\hat i + b\,\hat j + c\hat k:a,\,b,\,c\, \in \left\{ { - 1,\,1} \right\}} \right\}$$. Three non-coplanar vectors can be chosen from v in $${2^p}$$ ways. Then p is

Let $${S_n} = {\sum\limits_{k = 1}^{4n} {\left( { - 1} \right)} ^{{{k\left( {k + 1} \right)} \over 2}}}{k^2}.$$ Then $${S_n}$$can take value(s)

Perpendiculars are drawn from points on the line $\frac{x+2}{2}=\frac{y+1}{-1}=\frac{z}{3}$ to the plane $x+y+$ $z=3$. The foot of perpendiculars lie on the line

For $$a > b > c > 0,$$ the distance between $$(1, 1)$$ and the point of intersection of the lines $$ax + by + c = 0$$ and $$bx + ay + c = 0$$ is less than $$\left( {2\sqrt 2 } \right)$$. Then