Consider three planes :
`2x+py+6z=8, x+2y+qz=5 and x+y+3z=4`
Q. Three planes do not have any common point of intersection if :

Options

A. `p=2, q ne 3`
B. `p ne 2, q ne 3`
C. `p ne 2, q=3`
D. `p=2, q=3`

Answer

To determine the conditions under which the three planes do not have any common point of intersection, we can use the concept of the scalar triple product of the normal vectors of the planes. The planes given are:

1. \( 2x + py + 6z = 8 \)
2. \( x + 2y + qz = 5 \)
3. \( x + y + 3z = 4 \)

### Step 1: Identify the normal vectors of the planes
The normal vector of a plane \( Ax + By + Cz = D \) is given by the coefficients \( (A, B, C) \).

- For the first plane \( 2x + py + 6z = 8 \), the normal vector \( \mathbf{n_1} = (2, p, 6) \).
- For the second plane \( x + 2y + qz = 5 \), the normal vector \( \mathbf{n_2} = (1, 2, q) \).
- For the third plane \( x + y + 3z = 4 \), the normal vector \( \mathbf{n_3} = (1, 1, 3) \).

### Step 2: Set up the scalar triple product
The scalar triple product of the vectors \( \mathbf{n_1}, \mathbf{n_2}, \mathbf{n_3} \) can be represented as the determinant of the matrix formed by these vectors:

\[ \text{Det} = \begin{vmatrix} 2 & p & 6 \\ 1 & 2 & q \\ 1 & 1 & 3 \end{vmatrix} \]

### Step 3: Calculate the determinant
To find the determinant, we can expand it using the first row:

\[ \text{Det} = 2 \begin{vmatrix} 2 & q \\ 1 & 3 \end{vmatrix} - p \begin{vmatrix} 1 & q \\ 1 & 3 \end{vmatrix} + 6 \begin{vmatrix} 1 & 2 \\ 1 & 1 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. \( \begin{vmatrix} 2 & q \\ 1 & 3 \end{vmatrix} = (2 \cdot 3) - (q \cdot 1) = 6 - q \)

2. \( \begin{vmatrix} 1 & q \\ 1 & 3 \end{vmatrix} = (1 \cdot 3) - (q \cdot 1) = 3 - q \)

3. \( \begin{vmatrix} 1 & 2 \\ 1 & 1 \end{vmatrix} = (1 \cdot 1) - (2 \cdot 1) = 1 - 2 = -1 \)

Substituting these back into the determinant:

\[ \text{Det} = 2(6 - q) - p(3 - q) + 6(-1) \]
\[ = 12 - 2q - 3p + pq - 6 \]
\[ = pq - 3p - 2q + 6 \]

### Step 4: Set the condition for no common intersection
For the three planes to not have any common point of intersection, the scalar triple product must not equal zero:

\[ pq - 3p - 2q + 6 \neq 0 \]

### Step 5: Rearranging the condition
We can rearrange this condition to find the values of \( p \) and \( q \):

\[ pq - 3p - 2q + 6 \neq 0 \]

This inequality indicates that the planes do not intersect at a single point if the values of \( p \) and \( q \) do not satisfy this equation.

### Conclusion
The three planes do not have any common point of intersection if:

\[ p \neq 2 \quad \text{or} \quad q \neq 3 \]