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

To find the distance \( d \) from the point \( (3, 5, 0) \) to the plane described by the equation \( 6x + 9y - 7z + 14 = 0 \), you can use the formula: \[ d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \] where \( (x_0, y_0, z_0) \) is the point and \( Ax + By + Cz + D = 0 \) is the equation of the plane. Here, \( A = 6, B = 9, C = -7, D = 14 \), and \( (x_0, y_0, z_0) = (3, 5, 0) \). Plugging in the values: \[ d = \frac{|6(3) + 9(5) - 7(0) + 14|}{\sqrt{6^2 + 9^2 + (-7)^2}} \] Calculating the numerator: \[ = |18 + 45 + 0 + 14| = |77| = 77 \] Calculating the denominator: \[ = \sqrt{36 + 81 + 49} = \sqrt{166} \] Thus, the distance is: \[ d = \frac{77}{\sqrt{166}} \approx 5.94 \] So the distance \( d \) is \( \mathbf{5.94} \).