Step 1: Understanding the Concept:
To find the foot of the perpendicular from a point to a plane, we first find the equation of the line passing through the given point and perpendicular to the plane. The intersection of this line with the plane is the required foot.
Step 2: Key Formula or Approach:
1. The direction ratios of the normal to the plane $ax + by + cz + d = 0$ are $(a, b, c)$.
2. The equation of a line passing through $(x_1, y_1, z_1)$ with direction ratios $(a, b, c)$ is $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} = t$.
3. Any point on this line is $(x_1+at, y_1+bt, z_1+ct)$. Substitute this into the plane equation to find $t$.
Step 3: Detailed Explanation:
The given plane is $2x - 3y + z - 11 = 0$.
The normal vector to the plane has direction ratios $\langle 2, -3, 1 \rangle$.
The perpendicular line passes through $P(-1, 1, 2)$ and is parallel to the normal vector. Its symmetric equation is:
\[ \frac{x - (-1)}{2} = \frac{y - 1}{-3} = \frac{z - 2}{1} = t \]
Any general point on this line can be written in terms of $t$ as:
\[ x = 2t - 1 \]
\[ y = -3t + 1 \]
\[ z = t + 2 \]
Let this point be the foot of the perpendicular, $F(2t - 1, -3t + 1, t + 2)$.
Since $F$ lies on the plane, its coordinates must satisfy the plane equation:
\[ 2(2t - 1) - 3(-3t + 1) + (t + 2) - 11 = 0 \]
Expand and solve for $t$:
\[ 4t - 2 + 9t - 3 + t + 2 - 11 = 0 \]
\[ (4t + 9t + t) + (-2 - 3 + 2 - 11) = 0 \]
\[ 14t - 14 = 0 \]
\[ 14t = 14 \implies t = 1 \]
Now substitute $t = 1$ back into the coordinates of $F$:
\[ x = 2(1) - 1 = 1 \]
\[ y = -3(1) + 1 = -2 \]
\[ z = 1 + 2 = 3 \]
The coordinates of the foot of the perpendicular are $(1, -2, 3)$.
Step 4: Final Answer:
The coordinates are $(1, -2, 3)$.