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Let's analyze the problem and solve it step by step.

### Problem Summary:
You are asked to find:
1. The equation of the **tangent plane** to the surface \( x = 3y^2 + 3z^2 - 252 \) at the point \( (-6, -1, -9) \).
2. The equation of the **normal line** at the same point.

### Step 1: Tangent Plane Equation
The general form of the tangent plane to a surface defined implicitly as \( F(x, y, z) = 0 \) at a point \( (x_0, y_0, z_0) \) is given by:
\[
F_x(x_0, y_0, z_0)(x - x_0) + F_y(x_0, y_0, z_0)(y - y_0) + F_z(x_0, y_0, z_0)(z - z_0) = 0
\]
Where:
- \( F_x, F_y, F_z \) are partial derivatives of \( F(x, y, z) \) with respect to \( x, y, z \), respectively.

For the given surface, we can rewrite the equation as:
\[
F(y, z) = x - 3y^2 - 3z^2 + 252 = 0
\]
This is because the surface is given explicitly for \( x \).

#### Partial Derivatives:
- \( F_y = -6y \)
- \( F_z = -6z \)
- \( F_x = 1 \) (since \( x \) is explicitly given)

#### Evaluate at \( (-6, -1, -9) \):
- \( F_y(-1) = -6(-1) = 6 \)
- \( F_z(-9) = -6(-9) = 54 \)
- \( F_x = 1 \)

Thus, the equation of the tangent plane at \( (-6, -1, -9) \) is:
\[
1(x + 6) + 6(y + 1) + 54(z + 9) = 0
\]
Simplifying:
\[
x + 6 + 6(y + 1) + 54(z + 9) = 0
\]
\[
x + 6 + 6y + 6 + 54z + 486 = 0
\]
\[
x + 6y + 54z + 498 = 0
\]
Now, divide through by the coefficient of \( x \) (which is already 1) to satisfy the condition "make the coefficient of \( x \) equal to 1":
\[
x + 6y + 54z + 498 = 0
\]

### Step 2: Normal Line Equation
The normal line passes through the point \( (-6, -1, -9) \) and has a direction vector equal to the gradient of the surface at that point. The gradient is the vector of partial derivatives:
\[
\nabla F = \langle F_x, F_y, F_z \rangle = \langle 1, 6, 54 \rangle
\]
So the parametric equation of the normal line is:
\[
\langle x, y, z \rangle = \langle -6, -1, -9 \rangle + t \langle 1, 6, 54 \rangle
\]
This expands to:
\[
x = -6 + t(1)
\]
\[
y = -1 + t(6)
\]
\[
z = -9 + t(54)
\]

### Final Answer:
1. **Tangent Plane Equation**:
   \[
   x + 6y + 54z + 498 = 0
   \]
2. **Normal Line Equation**:
   \[
   \langle x, y, z \rangle = \langle -6, -1, -9 \rangle + t \langle 1, 6, 54 \rangle
   \]

Would you like more details on any of these steps? Here are some related questions you might want to explore:

1. What is the geometric interpretation of the tangent plane and normal line?
2. How are the partial derivatives used to find the normal vector?
3. How does the gradient relate to the direction of the normal line?
4. Why does the coefficient of \( x \) need to be set to 1 for the tangent plane equation?
5. What would the tangent plane and normal line look like graphically?

**Tip**: The gradient vector is always perpendicular (normal) to the surface at any given point, making it a crucial tool for both tangent planes and normal lines.

Find equations of the tangent plane and normal line to the surface x = 3y^2 + 3z^2 - 252 at the point (-6, -1, -9). Tangent Plane: (make the coefficient of x equal to 1).

This page details how to find the equations of the tangent plane and the normal line to the surface x = 3y^2 + 3z^2 - 252 at the point (-6, -1, -9). It covers multivariable calculus concepts including the use of partial derivatives and the gradient vector to derive these equations.

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