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To solve the problem, we need to find the equations of the tangent plane and the normal line to the surfaces at the given points \( P \) for each of the three parts.

### **Part (a)**: 
Surface equation: \( z^2 - 2x^2 - 2y^2 = 5 \) at point \( P(1, -1, 3) \).

1. **Find the gradient vector** \(\nabla f(x, y, z)\):
   \[
   f(x, y, z) = z^2 - 2x^2 - 2y^2 - 5
   \]
   \[
   \frac{\partial f}{\partial x} = -4x, \quad \frac{\partial f}{\partial y} = -4y, \quad \frac{\partial f}{\partial z} = 2z
   \]
   \[
   \nabla f(1, -1, 3) = (-4, 4, 6)
   \]

2. **Tangent plane equation**:
   The equation of the tangent plane at \( P(1, -1, 3) \) is given by:
   \[
   -4(x - 1) + 4(y + 1) + 6(z - 3) = 0
   \]
   Simplifying:
   \[
   -4x + 4y + 6z = 26
   \]

3. **Normal line equation**:
   The equation of the normal line passing through \( P(1, -1, 3) \) is:
   \[
   x = 1 - 4t, \quad y = -1 + 4t, \quad z = 3 + 6t
   \]

### **Part (b)**:
Surface equation: \( xyz = 6 \) at point \( P(1, 2, 3) \).

1. **Find the gradient vector** \(\nabla f(x, y, z)\):
   \[
   f(x, y, z) = xyz - 6
   \]
   \[
   \frac{\partial f}{\partial x} = yz, \quad \frac{\partial f}{\partial y} = xz, \quad \frac{\partial f}{\partial z} = xy
   \]
   \[
   \nabla f(1, 2, 3) = (6, 3, 2)
   \]

2. **Tangent plane equation**:
   The equation of the tangent plane at \( P(1, 2, 3) \) is given by:
   \[
   6(x - 1) + 3(y - 2) + 2(z - 3) = 0
   \]
   Simplifying:
   \[
   6x + 3y + 2z = 21
   \]

3. **Normal line equation**:
   The equation of the normal line passing through \( P(1, 2, 3) \) is:
   \[
   x = 1 + 6t, \quad y = 2 + 3t, \quad z = 3 + 2t
   \]

### **Part (c)**:
Surface equation: \( \sin(xz) - 4\cos(yz) = 4 \) at point \( P(\pi, \pi, 1) \).

1. **Find the gradient vector** \(\nabla f(x, y, z)\):
   \[
   f(x, y, z) = \sin(xz) - 4\cos(yz) - 4
   \]
   \[
   \frac{\partial f}{\partial x} = z\cos(xz), \quad \frac{\partial f}{\partial y} = 4z\sin(yz), \quad \frac{\partial f}{\partial z} = x\cos(xz) - 4y\sin(yz)
   \]
   Evaluate at \( P(\pi, \pi, 1) \):
   \[
   \nabla f(\pi, \pi, 1) = (\cos(\pi), 4\sin(\pi), \pi\cos(\pi) - 4\pi\sin(\pi)) = (-1, 0, -\pi)
   \]

2. **Tangent plane equation**:
   The equation of the tangent plane at \( P(\pi, \pi, 1) \) is:
   \[
   -1(x - \pi) + 0(y - \pi) - \pi(z - 1) = 0
   \]
   Simplifying:
   \[
   x + \pi z = \pi^2 + 1
   \]

3. **Normal line equation**:
   The equation of the normal line passing through \( P(\pi, \pi, 1) \) is:
   \[
   x = \pi - t, \quad y = \pi, \quad z = 1 - \frac{\pi}{2}t
   \]

Would you like to dive deeper into any specific part of these solutions? Here are some related questions you might find helpful:

1. How do you find the gradient of more complex surfaces?
2. What is the geometric interpretation of the tangent plane and normal line?
3. How would the equations change if the surface was not given implicitly?
4. Can the gradient ever be zero, and what does it mean in that case?
5. How do these methods extend to surfaces in higher dimensions?

**Tip:** Understanding the gradient vector's role is crucial, as it points in the direction of the steepest ascent and is perpendicular to the tangent plane at a point.

Learn how to solve tangent plane and normal line problems in multivariable calculus with detailed step-by-step solutions. This includes finding gradient vectors, equations of tangent planes, and normal lines at given points on implicit surfaces. Suitable for advanced undergraduate students.

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