If the lines x = ay + b, z = cy + d and x = a'z + b', y = c'z + d' are perpendicular, then :

The equation of the line passing through (–4, 3, 1), parallel

to the plane x + 2y – z – 5 = 0 and intersecting

the line $${{x + 1} \over { - 3}} = {{y - 3} \over 2} = {{z - 2} \over { - 1}}$$ is :

The plane through the intersection of the planes x + y + z = 1 and 2x + 3y – z + 4 = 0 and parallel to y-axis also passes through the point :

If the angle between the lines, $${x \over 2} = {y \over 2} = {z \over 1}$$

and $${{5 - x} \over { - 2}} = {{7y - 14} \over p} = {{z - 3} \over 4}\,\,$$ is $${\cos ^{ - 1}}\left( {{2 \over 3}} \right),$$ then p is equal to :