If the shortest distance between the straight lines $$3(x - 1) = 6(y - 2) = 2(z - 1)$$ and $$4(x - 2) = 2(y - \lambda ) = (z - 3),\lambda \in R$$ is $${1 \over {\sqrt {38} }}$$, then the integral value of $$\lambda$$ is equal to :

The lines x = ay $$-$$ 1 = z $$-$$ 2 and x = 3y $$-$$ 2 = bz $$-$$ 2, (ab $$\ne$$ 0) are coplanar, if :

Consider the line L given by the equation

$${{x - 3} \over 2} = {{y - 1} \over 1} = {{z - 2} \over 1}$$.

Let Q be the mirror image of the point (2, 3, $$-$$1) with respect to L. Let a plane P be such that it passes through Q, and the line L is perpendicular to P. Then which of the following points is on the plane P?

If the equation of plane passing through the mirror image of a point (2, 3, 1) with respect to line $${{x + 1} \over 2} = {{y - 3} \over 1} = {{z + 2} \over { - 1}}$$ and containing the line $${{x - 2} \over 3} = {{1 - y} \over 2} = {{z + 1} \over 1}$$ is $$\alpha$$x + $$\beta$$y + $$\gamma$$z = 24, then $$\alpha$$ + $$\beta$$ + $$\gamma$$ is equal to :