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 :

Let P be the plane, which contains the line of intersection of the planes, x + y + z – 6 = 0 and 2x + 3y + z + 5 = 0 and it is perpendicular to the xy-plane. Then the distance of the point (0, 0, 256) from P is equal to :

A plane passing through the points (0, –1, 0) and (0, 0, 1) and making an angle $${\pi \over 4}$$ with the plane y – z + 5 = 0, also passes through the point

If the line, $${{x - 1} \over 2} = {{y + 1} \over 3} = {{z - 2} \over 4}$$ meets the plane, x + 2y + 3z = 15 at a point P, then the distance of P from the origin is :