Let $$Q$$ be the foot of perpendicular drawn from the point $$P(1,2,3)$$ to the plane $$x+2 y+z=14$$. If $$R$$ is a point on the plane such that $$\angle P R Q=60^{\circ}$$, then the area of $$\triangle P Q R$$ is equal to :

If $$(2,3,9),(5,2,1),(1, \lambda, 8)$$ and $$(\lambda, 2,3)$$ are coplanar, then the product of all possible values of $$\lambda$$ is:

If the foot of the perpendicular from the point $$\mathrm{A}(-1,4,3)$$ on the plane $$\mathrm{P}: 2 x+\mathrm{m} y+\mathrm{n} z=4$$, is $$\left(-2, \frac{7}{2}, \frac{3}{2}\right)$$, then the distance of the point A from the plane P, measured parallel to a line with direction ratios $$3,-1,-4$$, is equal to :

Let the lines

$$\frac{x-1}{\lambda}=\frac{y-2}{1}=\frac{z-3}{2}$$ and

$$\frac{x+26}{-2}=\frac{y+18}{3}=\frac{z+28}{\lambda}$$ be coplanar

and $$\mathrm{P}$$ be the plane containing these two lines.

Then which of the following points does NOT lie on P?