The foot of the perpendicular drawn form the point `(4,2,3)` to the line joining the points `(1,-2,3) and (1,1,0)` lies on the plane:

Options

A. `x-y-2z=1`
B. `x-2y+z=1`
C. `2x+y-z=1`
D. `x+2y-z=1`

Answer

To find the foot of the perpendicular drawn from the point \( A(4, 2, 3) \) to the line joining the points \( B(1, -2, 3) \) and \( C(1, 1, 0) \), we can follow these steps:

### Step 1: Find the direction ratios of the line BC
The direction ratios of the line \( BC \) can be found using the coordinates of points \( B \) and \( C \).

\[ \text{Direction ratios of } BC = C - B = (1 - 1, 1 - (-2), 0 - 3) = (0, 3, -3) \]

### Step 2: Write the parametric equations of the line BC
Using the direction ratios, we can write the parametric equations of the line \( BC \):

Let \( \lambda \) be the parameter. The equations are:

\[ x = 1 + 0\lambda = 1 \]
\[ y = -2 + 3\lambda \]
\[ z = 3 - 3\lambda \]

### Step 3: Find the coordinates of a point D on the line BC
From the parametric equations, we can express the coordinates of point \( D \):

\[ D(1, -2 + 3\lambda, 3 - 3\lambda) \]

### Step 4: Find the direction ratios of AD
Now, we need to find the direction ratios of the line segment \( AD \):

\[ AD = D - A = (1 - 4, (-2 + 3\lambda) - 2, (3 - 3\lambda) - 3) = (-3, 3\lambda - 4, -3\lambda) \]

### Step 5: Set up the condition for perpendicularity
For \( AD \) to be perpendicular to \( BC \), the dot product of their direction ratios should be zero:

\[ (0, 3, -3) \cdot (-3, 3\lambda - 4, -3\lambda) = 0 \]

Calculating the dot product:

\[ 0 \cdot (-3) + 3 \cdot (3\lambda - 4) + (-3) \cdot (-3\lambda) = 0 \]
\[ 3(3\lambda - 4) + 9\lambda = 0 \]
\[ 9\lambda - 12 + 9\lambda = 0 \]
\[ 18\lambda - 12 = 0 \]
\[ 18\lambda = 12 \implies \lambda = \frac{12}{18} = \frac{2}{3} \]

### Step 6: Substitute \( \lambda \) back to find coordinates of D
Substituting \( \lambda = \frac{2}{3} \) into the equations for \( D \):

\[ D(1, -2 + 3 \cdot \frac{2}{3}, 3 - 3 \cdot \frac{2}{3}) = D(1, -2 + 2, 3 - 2) = D(1, 0, 1) \]

### Step 7: Find the equation of the plane
Now, we need to find the equation of the plane that contains point \( D(1, 0, 1) \) and is perpendicular to the line \( BC \). The normal vector to the plane can be taken as the direction ratios of \( BC \), which are \( (0, 3, -3) \).

The equation of the plane can be written as:

\[ 0(x - 1) + 3(y - 0) - 3(z - 1) = 0 \]
\[ 3y - 3z + 3 = 0 \implies y - z + 1 = 0 \]

### Conclusion
The equation of the plane is:

\[ y - z + 1 = 0 \]