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To solve this problem, we need to calculate the distance from the point \( P(3, 1, 1) \) to the plane passing through the points \( Q(1, -1, 2) \), \( R(0, -5, 6) \), and \( S(0, 3, 4) \). 

### Steps:

1. **Find the normal vector** to the plane:  
   We first need to find two vectors in the plane by subtracting the coordinates of the points.  
   \[
   \vec{QR} = R - Q = (0 - 1, -5 - (-1), 6 - 2) = (-1, -4, 4)
   \]
   \[
   \vec{QS} = S - Q = (0 - 1, 3 - (-1), 4 - 2) = (-1, 4, 2)
   \]

   Next, we take the cross product of these two vectors to get the normal vector \( \vec{n} \) to the plane:
   \[
   \vec{n} = \vec{QR} \times \vec{QS}
   \]
   Using the formula for the cross product:
   \[
   \vec{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & -4 & 4 \\ -1 & 4 & 2 \end{vmatrix}
   \]
   This results in:
   \[
   \vec{n} = \mathbf{i}(({-4})(2) - 4(4)) - \mathbf{j}((-1)(2) - 4(-1)) + \mathbf{k}((-1)(4) - (-4)(-1))
   \]
   Simplifying each component:
   \[
   \vec{n} = \mathbf{i}(-8 - 16) - \mathbf{j}(-2 + 4) + \mathbf{k}(-4 - 4) = -24\mathbf{i} - 2\mathbf{j} - 8\mathbf{k}
   \]
   So the normal vector is \( \vec{n} = (-24, -2, -8) \).

2. **Equation of the plane**:  
   The equation of a plane is given by:
   \[
   ax + by + cz = d
   \]
   Using the normal vector \( (-24, -2, -8) \), we substitute the coordinates of point \( Q(1, -1, 2) \) into the plane equation:
   \[
   -24(1) - 2(-1) - 8(2) = d
   \]
   \[
   -24 + 2 - 16 = d \quad \Rightarrow \quad d = -38
   \]
   Thus, the equation of the plane is:
   \[
   -24x - 2y - 8z = -38
   \]

3. **Distance from point \( P(3, 1, 1) \) to the plane**:  
   The distance from a point \( P(x_1, y_1, z_1) \) to a plane \( ax + by + cz = d \) is given by:
   \[
   D = \frac{|ax_1 + by_1 + cz_1 - d|}{\sqrt{a^2 + b^2 + c^2}}
   \]
   Substituting \( P(3, 1, 1) \) and the plane equation \( -24x - 2y - 8z = -38 \):
   \[
   D = \frac{|-24(3) - 2(1) - 8(1) - (-38)|}{\sqrt{(-24)^2 + (-2)^2 + (-8)^2}}
   \]
   \[
   D = \frac{|-72 - 2 - 8 + 38|}{\sqrt{576 + 4 + 64}}
   \]
   \[
   D = \frac{|-44|}{\sqrt{644}} = \frac{44}{\sqrt{644}} = \frac{44}{\sqrt{4 \times 161}} = \frac{44}{2\sqrt{161}} = \frac{22}{\sqrt{161}}
   \]

   Thus, the distance is \( \frac{22}{\sqrt{161}} \) or approximately 1.733.

Would you like any clarifications on the steps? Here are some related questions to further your understanding:

1. How do you find the equation of a plane given three points?
2. What are the properties of the cross product between two vectors?
3. How do you derive the formula for the distance from a point to a plane?
4. How can you verify that a point lies on a given plane?
5. What is the geometric interpretation of the normal vector to a plane?

**Tip**: When calculating the cross product, always verify the signs of each component to avoid errors in the normal vector.

Find the distance the point P(3, 1, 1) is to the plane through the three points Q(1, -1, 2), R(0, -5, 6), and S(0, 3, 4).

Learn how to calculate the distance from the point P(3, 1, 1) to the plane passing through points Q(1, -1, 2), R(0, -5, 6), and S(0, 3, 4) using vector algebra and the distance formula. This detailed solution is suitable for advanced high school students.

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