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The problem asks for the distance between a point \( P(-7, -2, 3) \) and a plane given by the equation \( 3x + 5y - 6z + 5 = 0 \).

The formula for the distance \( d \) from a point \( P(x_1, y_1, z_1) \) to a plane \( Ax + By + Cz + D = 0 \) is:

\[
d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}
\]

Here:
- \( A = 3 \), \( B = 5 \), \( C = -6 \), and \( D = 5 \) (from the plane equation),
- \( x_1 = -7 \), \( y_1 = -2 \), and \( z_1 = 3 \) (from the point coordinates).

### Step-by-step calculation:
1. Substitute the point coordinates and plane constants into the numerator:

\[
|A(-7) + B(-2) + C(3) + D| = |3(-7) + 5(-2) - 6(3) + 5|
\]
\[
= |-21 - 10 - 18 + 5| = |-44| = 44
\]

2. Calculate the denominator, which is the magnitude of the normal vector to the plane:

\[
\sqrt{A^2 + B^2 + C^2} = \sqrt{3^2 + 5^2 + (-6)^2} = \sqrt{9 + 25 + 36} = \sqrt{70}
\]

3. The distance \( d \) is:

\[
d = \frac{44}{\sqrt{70}} \approx 5.25
\]

Thus, the distance from the point \( P(-7, -2, 3) \) to the plane \( 3x + 5y - 6z + 5 = 0 \) is approximately 5.25 units.

Would you like further clarification on any step? 

Here are some related questions for deeper exploration:
1. How is the formula for distance to a plane derived from vector projections?
2. What is the geometric interpretation of the plane's normal vector in this problem?
3. How does changing the coordinates of the point \( P \) affect the distance?
4. What would happen to the distance if the plane equation changed to \( 3x + 5y - 6z = 0 \)?
5. How can you compute the distance between two parallel planes?

**Tip:** The numerator in the distance formula is the absolute value of the point-plane equation evaluated at the point, reflecting how far off the point is from satisfying the plane equation.

Të gjendet distanca e pikës P(-7, -2, 3) nga rrafshi: 3x + 5y - 6z + 5 = 0.

Calculate the distance from point P(-7, -2, 3) to the plane 3x + 5y - 6z + 5 = 0 using the distance formula for a point and a plane. This involves substituting the point and plane equation into the formula and simplifying to get an approximate distance of 5.25 units. Suitable for students in Grades 10-12.

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