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 :

The vertices B and C of a $$\Delta $$ABC lie on the line,

$${{x + 2} \over 3} = {{y - 1} \over 0} = {z \over 4}$$ such that BC = 5 units.

Then the area (in sq. units) of this triangle, given that the point A(1, –1, 2), is :