Show that the points (1, 1, 1) and (-3, 0, 1) are equidistant from the plane `3x+4y-12 z+13=0.`

Explanation

We know that the distance of the point `({x}_{1}, {y}_{1}, {z}_{1})` from the plane `{ax}+{by}+{cz}+` `{d}=0` is given by

`frac{|a x_{1}+b y_{1}+c z_{1}+d|}{sqrt{a^{2}+b^{2}+c^{2}}}`

Distance of the point `(1,1,1)` from the plane `3 {x}+4 {y}-12 {z}+13=0`

The required distance `=frac{|3(1)+4(1)-12(1)+13|}{sqrt{(3)^{2}+(4)^{2}+(-12)^{2}}}`


`=frac{|3+4-12+13|}{sqrt{9+16+144}} `

`=frac{8}{13} { units } quad-(1)`


...