Question

Find the perpendicular distance from point (1,-1,4) to the plane 2x-3y-2z=7

Answer 1

$$plane:2x-3y-2z-7=0$$

$$Point\ A(1, -1, 4)$$

The perpendicular distance from $Point\ A$ to the given plane is

$$d=\dfrac{|2(1)+(-3)(-1)+(-2)(4)-7|}{\sqrt{(2)^2+(-3)^2+(-2)^2}}=\dfrac{10}{\sqrt{17}}$$

$$d=\dfrac{10}{\sqrt{17}}$$