The foot of perpendicular from ` (-1, 2 , 3 ) ` on the plane passing through the points ` ( 1, - 1, 1 ) , ( 2, 1 , -2 ) ` and ` ( 3, - 1, - 1 ) ` is :

Options

A. ` (1, 1 , - 1 ) `
B. ` ( 2, 1, 2 ) `
C. ` ( - 2, 1 , 2 ) `
D. ` ( 2, - 2 , 1 ) `

Explanation

To find the foot of the perpendicular from the point \((-1, 2, 3)\) onto the plane defined by the points \((1, -1, 1)\), \((2, 1, -2)\), and \((3, -1, -1)\), we can follow these steps:

### Step 1: Find the normal vector of the plane
To find the equation of the plane, we first need to determine two vectors that lie on the plane. We can use the given points to create these vectors.

Let:
- \(A = (1, -1, 1)\)
- \(B = (2, 1, -2)\)
- \(C = (3, -1, -1)\)

Now, we can find the vectors \( \vec{AB} \) and \( \vec{AC} \):
\[ \vec{AB} = B - A = (2 - 1, 1 - (-1), -2 - 1) = (1, 2, -3) \]
\[ \vec{AC} = C - A = (3 - 1, -1 - (-1), -1 - 1) = (2, 0, -2) \]

Next, we find the normal vector \( \vec{n} \) to the plane by taking the cross product \( \vec{AB} \times \vec{AC} \):
\[ \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 2 & 0 & -2 \end{vmatrix} \]

Calculating the determinant:
\[ \vec{n} = \hat{i}(2 \cdot (-2) - 0 \cdot (-3)) - \hat{j}(1 \cdot (-2) - 2 \cdot (-3)) + \hat{k}(1 \cdot 0 - 2 \cdot 2) \]
\[ = \hat{i}(-4) - \hat{j}(-2 + 6) + \hat{k}(0 - 4) \]
\[ = -4\hat{i} - 4\hat{j} - 4\hat{k} \]
Thus, the normal vector \( \vec{n} = (-4, -4, -4) \) or simplified, \( (1, 1, 1) \).

### Step 2: Equation of the plane
Using point \( A(1, -1, 1) \) and the normal vector \( (1, 1, 1) \), the equation of the plane can be written as:
\[ 1(x - 1) + 1(y + 1) + 1(z - 1) = 0 \]
Simplifying this:
\[ x + y + z - 1 + 1 - 1 = 0 \implies x + y + z = 1 \]

### Step 3: Find the foot of the perpendicular
The foot of the perpendicular from the point \((-1, 2, 3)\) to the plane can be found using the parametric equations of the line passing through \((-1, 2, 3)\) in the direction of the normal vector \((1, 1, 1)\):
\[ x = -1 + t, \quad y = 2 + t, \quad z = 3 + t \]

### Step 4: Substitute into the plane equation
Substituting these parametric equations into the plane equation \(x + y + z = 1\):
\[ (-1 + t) + (2 + t) + (3 + t) = 1 \]
\[ -1 + t + 2 + t + 3 + t = 1 \]
\[ 3 + 3t = 1 \implies 3t = 1 - 3 \implies 3t = -2 \implies t = -\frac{2}{3} \]

### Step 5: Find the coordinates of the foot of the perpendicular
Now substituting \(t = -\frac{2}{3}\) back into the parametric equations:
\[ x = -1 - \frac{2}{3} = -\frac{3}{3} - \frac{2}{3} = -\frac{5}{3} \]
\[ y = 2 - \frac{2}{3} = \frac{6}{3} - \frac{2}{3} = \frac{4}{3} \]
\[ z = 3 - \frac{2}{3} = \frac{9}{3} - \frac{2}{3} = \frac{7}{3} \]

Thus, the foot of the perpendicular is:
\[ \left(-\frac{5}{3}, \frac{4}{3}, \frac{7}{3}\right) \]