Consider a tetrahedron `D-ABC` with position vectors if its angular points as
A(1, 1, 1), B(1, 2, 3), C(1, 1, 2)
and centre of tetrahedron `((3)/(2), (3)/(4),2)`.
Q. If N be the foot of the perpendicular from point D on the plane face ABC then the position vector of N are :

Options

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

Answer

To find the position vector of point N, the foot of the perpendicular from point D to the plane defined by triangle ABC, we will follow these steps:

### Step 1: Determine the coordinates of point D
We know the centroid (center) of the tetrahedron D-ABC is given as \( \left( \frac{3}{2}, \frac{3}{4}, 2 \right) \). The coordinates of points A, B, and C are:
- A(1, 1, 1)
- B(1, 2, 3)
- C(1, 1, 2)

Let the coordinates of point D be \( (x, y, z) \). The formula for the centroid of a tetrahedron with vertices at \( A, B, C, D \) is:

\[ \text{Centroid} = \left( \frac{x_A + x_B + x_C + x_D}{4}, \frac{y_A + y_B + y_C + y_D}{4}, \frac{z_A + z_B + z_C + z_D}{4} \right) \]

Substituting the known values:

\[ \frac{1 + 1 + 1 + x}{4} = \frac{3}{2} \quad \Rightarrow \quad 1 + 1 + 1 + x = 6 \quad \Rightarrow \quad x = 3 \]

\[ \frac{1 + 2 + 1 + y}{4} = \frac{3}{4} \quad \Rightarrow \quad 4 + y = 3 \quad \Rightarrow \quad y = -1 \]

\[ \frac{1 + 3 + 2 + z}{4} = 2 \quad \Rightarrow \quad 6 + z = 8 \quad \Rightarrow \quad z = 2 \]

Thus, the coordinates of point D are \( D(3, -1, 2) \).

### Step 2: Find the equation of the plane ABC
To find the equation of the plane defined by points A, B, and C, we can use the normal vector method. The points are:
- A(1, 1, 1)
- B(1, 2, 3)
- C(1, 1, 2)

We can find two vectors in the plane:
- \( \vec{AB} = B - A = (1, 2, 3) - (1, 1, 1) = (0, 1, 2) \)
- \( \vec{AC} = C - A = (1, 1, 2) - (1, 1, 1) = (0, 0, 1) \)

The normal vector \( \vec{n} \) to the plane can be found using the cross product \( \vec{AB} \times \vec{AC} \):

\[ \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{vmatrix} = \hat{i}(1 \cdot 1 - 2 \cdot 0) - \hat{j}(0 \cdot 1 - 2 \cdot 0) + \hat{k}(0 \cdot 0 - 0 \cdot 1) = \hat{i}(1) + \hat{j}(0) + \hat{k}(0) = (1, 0, 0) \]

The equation of the plane can be written as:

\[ 1(x - 1) + 0(y - 1) + 0(z - 1) = 0 \quad \Rightarrow \quad x - 1 = 0 \quad \Rightarrow \quad x = 1 \]

### Step 3: Find the foot of the perpendicular N from point D to the plane
The coordinates of point N must satisfy the plane equation \( x = 1 \). Let the coordinates of N be \( (1, y_N, z_N) \).

The direction ratios of the line DN from D to N are:

\[ (1 - 3, y_N - (-1), z_N - 2) = (-2, y_N + 1, z_N - 2) \]

Since the plane's normal vector is \( (1, 0, 0) \), the direction ratios of DN must be perpendicular to the normal vector. Therefore, we set up the dot product:

\[ (-2, y_N + 1, z_N - 2) \cdot (1, 0, 0) = 0 \quad \Rightarrow \quad -2 = 0 \]

This condition is satisfied for any values of \( y_N \) and \( z_N \). Thus, we can choose \( y_N = -1 \) and \( z_N = 2 \).

### Final Result
The coordinates of point N are \( N(1, -1, 2) \).

### Summary
The position vector of N is \( \mathbf{N} = (1, -1, 2) \).

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