The foot of perpendicular from the origin `O` to a plane `P` which meets the co-ordinate axes at the point `A,B,C` is `(2,a,4),a in N`.If the volume of the tetrahedron `OABC` is `144` `unit ^(3)`,then which of the following points is `NOT` on `P`?

Options

A. `(0,4,4)`
B. `(3,0,4)`
C. ` (0,6,3)`
D. ` (2,2,4)`

Answer

To solve the problem step by step, we will follow the reasoning provided in the video transcript and derive the necessary equations and calculations.

### Step 1: Identify the coordinates of the foot of the perpendicular
The foot of the perpendicular from the origin \( O(0, 0, 0) \) to the plane \( P \) is given as \( (2, a, 4) \), where \( a \) is a natural number.

### Step 2: Write the equation of the plane
The general equation of a plane that passes through the point \( (2, a, 4) \) can be expressed in the form:
\[ 2(x - 2) + a(y - a) + 4(z - 4) = 0 \]
Expanding this gives:
\[ 2x - 4 + ay - a^2 + 4z - 16 = 0 \]
Rearranging, we have:
\[ 2x + ay + 4z = a^2 + 20 \]

### Step 3: Volume of the tetrahedron \( OABC \)
The volume \( V \) of the tetrahedron formed by the points \( O(0, 0, 0) \), \( A(2, 0, 0) \), \( B(0, a, 0) \), and \( C(0, 0, 4) \) is given by the formula:
\[ V = \frac{1}{6} \times \text{Base Area} \times \text{Height} \]
The volume can also be calculated using the formula:
\[ V = \frac{1}{6} \times \text{det} \begin{pmatrix} x_1 & y_1 & z_1 & 1 \\ x_2 & y_2 & z_2 & 1 \\ x_3 & y_3 & z_3 & 1 \\ x_4 & y_4 & z_4 & 1 \end{pmatrix} \]
For our points:
\[ V = \frac{1}{6} \times \left| \begin{vmatrix} 2 & 0 & 0 & 1 \\ 0 & a & 0 & 1 \\ 0 & 0 & 4 & 1 \\ 0 & 0 & 0 & 1 \end{vmatrix} \right| \]
Calculating the determinant:
\[ = \frac{1}{6} \times (2 \cdot a \cdot 4) = \frac{8a}{6} = \frac{4a}{3} \]
Setting this equal to the given volume of 144:
\[ \frac{4a}{3} = 144 \]
Multiplying both sides by 3:
\[ 4a = 432 \]
Dividing by 4:
\[ a = 108 \]

### Step 4: Substitute \( a \) back into the plane equation
Now substituting \( a = 108 \) into the plane equation:
\[ 2x + 108y + 4z = 108^2 + 20 \]
Calculating \( 108^2 \):
\[ 108^2 = 11664 \]
So the equation becomes:
\[ 2x + 108y + 4z = 11684 \]

### Step 5: Check which points lie on the plane
We need to check which of the given points does not satisfy the plane equation \( 2x + 108y + 4z = 11684 \).

1. **Point \( (0, 4, 4) \)**:
\[ 2(0) + 108(4) + 4(4) = 0 + 432 + 16 = 448 \quad \text{(not on the plane)} \]

2. **Point \( (3, 0, 4) \)**:
\[ 2(3) + 108(0) + 4(4) = 6 + 0 + 16 = 22 \quad \text{(not on the plane)} \]

3. **Point \( (0, 6, 3) \)**:
\[ 2(0) + 108(6) + 4(3) = 0 + 648 + 12 = 660 \quad \text{(not on the plane)} \]

4. **Point \( (2, 2, 4) \)**:
\[ 2(2) + 108(2) + 4(4) = 4 + 216 + 16 = 236 \quad \text{(not on the plane)} \]

### Conclusion
The point that does not lie on the plane \( P \) is \( (3, 0, 4) \).