Find the distance between the point $$ (3,5,0) $$ and the plane $$ 6 x+9 y-7 z+14=0 $$ $$ d= $$ Ex: $$ 5.67^{6} $$

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To find the distance between a point and a plane, we can use the formula:

$$ d = \frac{|a(x_1) + b(y_1) + c(z_1) + d|}{\sqrt{a^2 + b^2 + c^2}} $$

where:
- $$ (x_1, y_1, z_1) $$ are the coordinates of the point,
- $$ a, b, c, d $$ are the coefficients of the plane equation.

Given point: $$ (3, 5, 0) $$
Given plane equation: $$ 6x + 9y - 7z + 14 = 0 $$

Substitute the values into the formula:

$$ d = \frac{|6(3) + 9(5) - 7(0) + 14|}{\sqrt{6^2 + 9^2 + (-7)^2}} $$

$$ d = \frac{|18 + 45 + 14|}{\sqrt{36 + 81 + 49}} $$

$$ d = \frac{|77|}{\sqrt{166}} $$

$$ d = \frac{77}{\sqrt{166}} $$

Now, we can calculate the value of $$ d $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{77}{\sqrt{166}}$$
- step1: Multiply by the Conjugate:
  $$\frac{77\sqrt{166}}{\sqrt{166}\times \sqrt{166}}$$
- step2: Multiply the numbers:
  $$\frac{77\sqrt{166}}{166}$$
The distance between the point $$ (3, 5, 0) $$ and the plane $$ 6x + 9y - 7z + 14 = 0 $$ is approximately $$ 5.976359 $$.
The distance between the point $$ (3, 5, 0) $$ and the plane $$ 6x + 9y - 7z + 14 = 0 $$ is approximately $$ 5.98 $$.

Find the distance between the point and the plane

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