If the points (1, 1, p) and (–3, 0, 1) be equidistant from the plane , then find the value of p.

The distance of a point with position vector from the plane is .

The position vector of point (1,1,p) is given as -

⇒

The position vector of point (-3,0,1) is given as -

⇒

It is given that the points (1,1,p) and (-3,0,1) are equidistant from the plane

∴

⇒ 20 - 12p = 8

⇒ 20 - 12p = 8 or, 20 - 12p = -8

⇒ 12p = 12 or, 12p = 28

∴ p = 1 or, p = 7/3

us, the possible values of p are 1 and 7/3.