Find the equation of the plane containing the points P(2, 3, -1), Q(6, -1, 0) and R(3, 1, 4).

Answer

To find the equation of the plane containing the points P(2, 3, -1), Q(6, -1, 0), and R(3, 1, 4), we will follow these steps:

### Step 1: Determine the vectors PQ and PR
First, we need to find the vectors PQ and PR using the coordinates of the points.

- **Vector PQ**:
\[ \vec{PQ} = Q - P = (6 - 2, -1 - 3, 0 - (-1)) = (4, -4, 1) \]

- **Vector PR**:
\[ \vec{PR} = R - P = (3 - 2, 1 - 3, 4 - (-1)) = (1, -2, 5) \]

### Step 2: Find the normal vector to the plane
The normal vector \(\vec{n}\) to the plane can be found by taking the cross product of vectors \(\vec{PQ}\) and \(\vec{PR}\).

\[ \vec{n} = \vec{PQ} \times \vec{PR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -4 & 1 \\ 1 & -2 & 5 \end{vmatrix} \]

Calculating the determinant:

\[ \vec{n} = \hat{i}((-4)(5) - (1)(-2)) - \hat{j}((4)(5) - (1)(1)) + \hat{k}((4)(-2) - (-4)(1)) \]

\[ = \hat{i}(-20 + 2) - \hat{j}(20 - 1) + \hat{k}(-8 + 4) \]

\[ = \hat{i}(-18) - \hat{j}(19) + \hat{k}(-4) \]

Thus, the normal vector is:
\[ \vec{n} = (-18, -19, -4) \]

### Step 3: Use the point-normal form of the plane equation
The equation of the plane can be expressed in the form:
\[ \vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n} \]
where \(\vec{a}\) is a point on the plane (we can use point P).

Substituting \(\vec{a} = (2, 3, -1)\) and \(\vec{n} = (-18, -19, -4)\):

Calculating \(\vec{a} \cdot \vec{n}\):
\[ \vec{a} \cdot \vec{n} = 2(-18) + 3(-19) + (-1)(-4) = -36 - 57 + 4 = -89 \]

Thus, the equation of the plane is:
\[ \vec{r} \cdot (-18, -19, -4) = -89 \]

### Step 4: Convert to Cartesian form
Let \(\vec{r} = (x, y, z)\), then:
\[ -18x - 19y - 4z = -89 \]
Rearranging gives:
\[ 18x + 19y + 4z = 89 \]

### Final Answer
The equation of the plane containing the points P, Q, and R is:
\[ 18x + 19y + 4z = 89 \]

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