To find the foot of the perpendicular from the point \((-1, 2, 3)\) to the plane defined by the points \((1, -1, 1)\), \((2, 1, -2)\), and \((3, -1, 1)\), we will follow these steps:
### Step 1: Find the normal vector of the plane
To find the equation of the plane, we need to determine two vectors that lie in the plane. We can form these vectors using the given points.
Let:
- \( A = (1, -1, 1) \)
- \( B = (2, 1, -2) \)
- \( C = (3, -1, 1) \)
We can find vectors \( \vec{AB} \) and \( \vec{AC} \):
\[
\vec{AB} = B - A = (2 - 1, 1 - (-1), -2 - 1) = (1, 2, -3)
\]
\[
\vec{AC} = C - A = (3 - 1, -1 - (-1), 1 - 1) = (2, 0, 0)
\]
Next, we find the normal vector \( \vec{n} \) to the plane by taking the cross product \( \vec{AB} \times \vec{AC} \):
\[
\vec{n} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
1 & 2 & -3 \\
2 & 0 & 0
\end{vmatrix}
\]
Calculating the determinant:
\[
\vec{n} = \hat{i}(2 \cdot 0 - (-3) \cdot 0) - \hat{j}(1 \cdot 0 - (-3) \cdot 2) + \hat{k}(1 \cdot 0 - 2 \cdot 2)
\]
\[
= 0\hat{i} - (-6)\hat{j} - 4\hat{k} = (0, 6, -4)
\]
Thus, the normal vector is \( \vec{n} = (0, 6, -4) \).
### Step 2: Write the equation of the plane
Using the normal vector \( (0, 6, -4) \) and point \( A(1, -1, 1) \), the equation of the plane can be written as:
\[
0(x - 1) + 6(y + 1) - 4(z - 1) = 0
\]
Simplifying this gives:
\[
6y - 4z + 6 + 4 = 0 \implies 6y - 4z + 10 = 0
\]
This can be rearranged to:
\[
3y - 2z + 5 = 0
\]
### 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 direction ratios of the normal vector. The parametric equations of the line through the point \((-1, 2, 3)\) in the direction of the normal vector \( (0, 6, -4) \) are:
\[
x = -1 + 0t = -1
\]
\[
y = 2 + 6t
\]
\[
z = 3 - 4t
\]
### Step 4: Substitute into the plane equation
Substituting \( y \) and \( z \) into the plane equation \( 3y - 2z + 5 = 0 \):
\[
3(2 + 6t) - 2(3 - 4t) + 5 = 0
\]
Expanding this gives:
\[
6 + 18t - 6 + 8t + 5 = 0
\]
\[
26t + 5 = 0 \implies t = -\frac{5}{26}
\]
### Step 5: Find the coordinates of the foot of the perpendicular
Substituting \( t = -\frac{5}{26} \) back into the parametric equations:
\[
x = -1
\]
\[
y = 2 + 6\left(-\frac{5}{26}\right) = 2 - \frac{30}{26} = 2 - \frac{15}{13} = \frac{26}{13} - \frac{15}{13} = \frac{11}{13}
\]
\[
z = 3 - 4\left(-\frac{5}{26}\right) = 3 + \frac{20}{26} = 3 + \frac{10}{13} = \frac{39}{13} + \frac{10}{13} = \frac{49}{13}
\]
Thus, the coordinates of the foot of the perpendicular are:
\[
\left(-1, \frac{11}{13}, \frac{49}{13}\right)
\]
### Final Answer
The foot of the perpendicular from the point \((-1, 2, 3)\) to the plane is:
\[
\left(-1, \frac{11}{13}, \frac{49}{13}\right)
\]
