If the points (1, 1, ) and (-3, 0, 1) are equidistant from the plane r (3 i + 4 j - 12 k ) + 13 = 0 , then the value of are

Step-by-step solution

The equation of the given plane in Cartesian form is 3x + 4y - 12z + 13 = 0 . The perpendicular distance of a point (x₁, y₁, z₁) from the plane Ax + By + Cz + D = 0 is given by |Ax₁ + By₁ + Cz₁ + D| A^2 + B^2 + C^2 . The distance of the point (1, 1, ) from the plane is: d₁ = |3(1) + 4(1) - 12( ) + 13| 3^2 + 4^2 + (-12)^2 = |20 - 12 | 13 The distance of the point (-3, 0, 1) from the plane is: d₂ = |3(-3) + 4(0) - 12(1) + 13| 3^2 + 4^2 + (-12)^2 = |-9 - 12 + 13| 13 = |-8| 13 = 8 13 Since the points are equidistant fr