The area (in sq. units) of an equilateral triangle inscribed in the parabola y² = 8x, with one of its vertices on the vertex of this parabola, is :

Let f : (–1, $$\infty $$) $$ \to $$ R be defined by f(0) = 1 and
f(x) = $${1 \over x}{\log _e}\left( {1 + x} \right)$$, x $$ \ne $$ 0. Then the function f :

Let A = {X = (x, y, z)^T: PX = 0 and

x² + y² + z² = 1} where

$$P = \left[ {\matrix{ 1 & 2 & 1 \cr { - 2} & 3 & { - 4} \cr 1 & 9 & { - 1} \cr } } \right]$$,

then the set A :

Let the position vectors of points 'A' and 'B' be
$$\widehat i + \widehat j + \widehat k$$ and $$2\widehat i + \widehat j + 3\widehat k$$, respectively. A point 'P' divides the line segment AB internally in the ratio $$\lambda $$ : 1 ( $$\lambda $$ > 0). If O is the origin and
$$\overrightarrow {OB} .\overrightarrow {OP} - 3{\left| {\overrightarrow {OA} \times \overrightarrow {OP} } \right|^2} = 6$$, then $$\lambda $$ is equal to______.