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?

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 :

Let the plane ax + by + cz = d pass through (2, 3, $$-$$5) and is perpendicular to the planes
2x + y $$-$$ 5z = 10 and 3x + 5y $$-$$ 7z = 12. If a, b, c, d are integers d > 0 and gcd (|a|, |b|, |c|, d) = 1, then the value of a + 7b + c + 20d is equal to :