Point, Plane: $(- 6, 0, 0), 2 x - 3 y + 6 z - 2 = 0$ .

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Solution

We know that the distance between a point $p (x_{1}, y_{1}, z_{1})$ and a plane $A x + B y + C z = D$ is given by,
$d = ∣ ∣ ∣ ∣ \frac{A x_{1} + B y_{1} + C z_{1} - D}{\sqrt{A^{2} + B^{2} + C^{2}}} ∣ ∣ ∣ ∣ . . . . (1)$
Thus distance of point $(- 6, 0, 0)$ from plane $2 x - 3 y + 6 z - 2 = 0$ is
$d = ∣ ∣ ∣ ∣ ∣ \frac{2 (- 6) - 3 \times 0 + 6 \times 0 - 2}{\sqrt{{(1)}^{2} + {(2)}^{2} + {(- 2)}^{2}}} ∣ ∣ ∣ ∣ ∣ = ∣ ∣ ∣ \frac{- 14}{\sqrt{49}} ∣ ∣ ∣ = \frac{14}{7} = 2$

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Point, Plane: (−6,0,0),2x−3y+6z−2=0.

Point, Plane: $(- 6, 0, 0), 2 x - 3 y + 6 z - 2 = 0$ .