Let two vertices of a triangle ABC be (2, 4, 6) and (0, $$-$$2, $$-$$5), and its centroid be (2, 1, $$-$$1). If the image of the third vertex in the plane $$x+2y+4z=11$$ is $$(\alpha,\beta,\gamma)$$, then $$\alpha\beta+\beta\gamma+\gamma\alpha$$ is equal to :

Let O be the origin and the position vector of the point P be $$ - \widehat i - 2\widehat j + 3\widehat k$$. If the position vectors of the points A, B and C are $$ - 2\widehat i + \widehat j - 3\widehat k,2\widehat i + 4\widehat j - 2\widehat k$$ and $$ - 4\widehat i + 2\widehat j - \widehat k$$ respectively, then the projection of the vector $$\overrightarrow {OP} $$ on a vector perpendicular to the vectors $$\overrightarrow {AB} $$ and $$\overrightarrow {AC} $$ is :

The negation of the statement $$(p \vee q) \wedge (q \vee ( \sim r))$$ is :

Let P be the point of intersection of the line $${{x + 3} \over 3} = {{y + 2} \over 1} = {{1 - z} \over 2}$$ and the plane $$x+y+z=2$$. If the distance of the point P from the plane $$3x - 4y + 12z = 32$$ is q, then q and 2q are the roots of the equation :