If the plane 2x – y + 2z + 3 = 0 has the distances $${1 \over 3}$$ and $${2 \over 3}$$ units from the planes 4x – 2y + 4z + $$\lambda $$ = 0 and 2x – y + 2z + $$\mu $$ = 0, respectively, then the maximum value of $$\lambda $$ + $$\mu $$ is equal to :

A perpendicular is drawn from a point on the line $${{x - 1} \over 2} = {{y + 1} \over { - 1}} = {z \over 1}$$ to the plane x + y + z = 3 such that the foot of the perpendicular Q also lies on the plane x – y + z = 3. Then the co-ordinates of Q are :

If the length of the perpendicular from the point ($$\beta $$, 0, $$\beta $$) ($$\beta $$ $$ \ne $$ 0) to the line,
$${x \over 1} = {{y - 1} \over 0} = {{z + 1} \over { - 1}}$$ is $$\sqrt {{3 \over 2}} $$, then $$\beta $$ is equal to :

If Q(0, –1, –3) is the image of the point P in the plane 3x – y + 4z = 2 and R is the point (3, –1, –2), then the area (in sq. units) of $$\Delta $$PQR is :