Let $${{x - 2} \over 3} = {{y + 1} \over { - 2}} = {{z + 3} \over { - 1}}$$ lie on the plane $$px - qy + z = 5$$, for some p, q $$\in$$ R. The shortest distance of the plane from the origin is :
Let Q be the mirror image of the point P(1, 2, 1) with respect to the plane x + 2y + 2z = 16. Let T be a plane passing through the point Q and contains the line $$\overrightarrow r = - \widehat k + \lambda \left( {\widehat i + \widehat j + 2\widehat k} \right),\,\lambda \in R$$. Then, which of the following points lies on T ?
$$\overrightarrow a = \widehat i + 4\widehat j + 3\widehat k$$
$$\overrightarrow b = 2\widehat i + \alpha \widehat j + 4\widehat k,\,\alpha \in R$$
$$\overrightarrow c = 3\widehat i - 2\widehat j + 5\widehat k$$
If $$\alpha$$ is the smallest positive integer for which $$\overrightarrow a ,\,\overrightarrow b ,\,\overrightarrow c $$ are noncollinear, then the length of the median, in $$\Delta$$ABC, through A is :
If the mirror image of the point (2, 4, 7) in the plane 3x $$-$$ y + 4z = 2 is (a, b, c), then 2a + b + 2c is equal to: