The distance of the point $(2,-1,0)$ from the plane $2 x+\mathrm{y}+2 \mathrm{z}+8=0$ is

The distance of the point $(2,-1,0)$ from the plane $2 x+\mathrm{y}+2 \mathrm{z}+8=0$ is
  1. $\frac{17}{3}$ units
  2. $\frac{13}{3}$ units
  3. $\frac{7}{3}$ units
  4. $\frac{11}{3}$ units

Solution

$\mathrm{d}=\left|\frac{2(2)+(-1)(1)+0+8}{\sqrt{4+1+4}}\right|=\frac{11}{3}$ units

Asked in: MHT CET 2020 (15 Oct Shift 2)

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