Find the vector equation of the point determined by the points A (3,-1,2), B (5, 2, 4) and C (-1, -1, 6) Hence, find the distance of the plane, thus obtained, from the origin.

Answer

To find the vector equation of the plane determined by the points A(3, -1, 2), B(5, 2, 4), and C(-1, -1, 6), and then to calculate the distance of this plane from the origin, we can follow these steps:

### Step 1: Find the position vectors of points A, B, and C.
Let:
- \(\vec{A} = 3\hat{i} - 1\hat{j} + 2\hat{k}\)
- \(\vec{B} = 5\hat{i} + 2\hat{j} + 4\hat{k}\)
- \(\vec{C} = -1\hat{i} - 1\hat{j} + 6\hat{k}\)

### Step 2: Find the vectors \(\vec{AB}\) and \(\vec{AC}\).
\[ \vec{AB} = \vec{B} - \vec{A} = (5 - 3)\hat{i} + (2 + 1)\hat{j} + (4 - 2)\hat{k} = 2\hat{i} + 3\hat{j} + 2\hat{k} \]
\[ \vec{AC} = \vec{C} - \vec{A} = (-1 - 3)\hat{i} + (-1 + 1)\hat{j} + (6 - 2)\hat{k} = -4\hat{i} + 0\hat{j} + 4\hat{k} \]

### Step 3: Find the normal vector \(\vec{n}\) to the plane using the cross product \(\vec{AB} \times \vec{AC}\).
\[ \vec{n} = \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 2 \\ -4 & 0 & 4 \end{vmatrix} \]

Calculating the determinant:
\[ \vec{n} = \hat{i}(3 \cdot 4 - 2 \cdot 0) - \hat{j}(2 \cdot 4 - 2 \cdot -4) + \hat{k}(2 \cdot 0 - 3 \cdot -4) \]
\[ = \hat{i}(12) - \hat{j}(8 + 8) + \hat{k}(0 + 12) \]
\[ = 12\hat{i} - 16\hat{j} + 12\hat{k} \]

### Step 4: Write the equation of the plane.
The general form of the plane equation is given by:
\[ \vec{r} \cdot \vec{n} = d \]
Substituting \(\vec{n}\) and using point A to find \(d\):
\[ \vec{r} \cdot (12\hat{i} - 16\hat{j} + 12\hat{k}) = d \]
Using point A(3, -1, 2):
\[ d = 12 \cdot 3 - 16 \cdot (-1) + 12 \cdot 2 = 36 + 16 + 24 = 76 \]
Thus, the equation of the plane is:
\[ 12x - 16y + 12z = 76 \]

### Step 5: Simplify the equation of the plane.
Dividing the entire equation by 4:
\[ 3x - 4y + 3z = 19 \]

### Step 6: Calculate the distance from the origin to the plane.
The formula for the distance \(D\) from a point \((x_1, y_1, z_1)\) to the plane \(Ax + By + Cz = D\) is given by:
\[ D = \frac{|Ax_1 + By_1 + Cz_1 - D|}{\sqrt{A^2 + B^2 + C^2}} \]
For the origin \((0, 0, 0)\):
\[ D = \frac{|3 \cdot 0 - 4 \cdot 0 + 3 \cdot 0 - 19|}{\sqrt{3^2 + (-4)^2 + 3^2}} \]
\[ = \frac{|-19|}{\sqrt{9 + 16 + 9}} = \frac{19}{\sqrt{34}} \]

### Final Answer:
The distance of the plane from the origin is \(\frac{19}{\sqrt{34}}\).