The value of the` lambda` so that P, Q, R, S on the sides OA, OB, OC and AB of a regular tetrahedron are coplanar. When `(OP)/(OA)=1/3 ;(OQ)/(OB)=1/2` and `(OS)/(AB)=lambda` is (A) `lamda=1/2` (B) `lamda=-1` (C) `lamda=0` (D) `lamda=2`

Options

A. `lamda = (1)/(2)`
B. `lamda =-1`
C. `lamda =0`
D. for no value of `lamda`

Explanation

To solve the problem of finding the value of \( \lambda \) such that points \( P, Q, R, S \) on the sides \( OA, OB, OC, \) and \( AB \) of a regular tetrahedron are coplanar, we can follow these steps:

### Step 1: Define the Position Vectors
Let the position vectors of points \( A, B, C, \) and \( O \) be represented as:
- \( \vec{A} \)
- \( \vec{B} \)
- \( \vec{C} \)
- \( \vec{O} \)

### Step 2: Express Points \( P, Q, R, S \)
Given the ratios:
- \( \frac{OP}{OA} = \frac{1}{3} \) implies \( \vec{P} = \frac{1}{3} \vec{A} \)
- \( \frac{OQ}{OB} = \frac{1}{2} \) implies \( \vec{Q} = \frac{1}{2} \vec{B} \)
- \( \frac{OS}{AB} = \lambda \) implies \( \vec{S} = \lambda (\vec{B} - \vec{A}) + \vec{A} \)

### Step 3: Express Point \( R \)
Since \( R \) lies on \( OC \), we can express it as:
- \( \vec{R} = t \vec{C} \) for some \( t \) such that \( 0 \leq t \leq 1 \).

### Step 4: Use the Condition for Coplanarity
The points \( P, Q, R, S \) are coplanar if the scalar triple product of the vectors \( \vec{PQ}, \vec{PR}, \vec{PS} \) is zero.

1. **Find \( \vec{PQ} \)**:
\[ \vec{PQ} = \vec{Q} - \vec{P} = \left(\frac{1}{2} \vec{B} - \frac{1}{3} \vec{A}\right) \]

2. **Find \( \vec{PR} \)**:
\[ \vec{PR} = \vec{R} - \vec{P} = t \vec{C} - \frac{1}{3} \vec{A} \]

3. **Find \( \vec{PS} \)**:
\[ \vec{PS} = \vec{S} - \vec{P} = \left(\lambda (\vec{B} - \vec{A}) + \vec{A}\right) - \frac{1}{3} \vec{A} = \lambda \vec{B} + \left(1 - \frac{1}{3}\right) \vec{A} = \lambda \vec{B} + \frac{2}{3} \vec{A} \]

### Step 5: Set Up the Determinant for Coplanarity
The points are coplanar if:
\[ \begin{vmatrix} \vec{PQ} & \vec{PR} & \vec{PS} \end{vmatrix} = 0 \]

Substituting the vectors:
\[ \begin{vmatrix} \left(\frac{1}{2} \vec{B} - \frac{1}{3} \vec{A}\right) & \left(t \vec{C} - \frac{1}{3} \vec{A}\right) & \left(\lambda \vec{B} + \frac{2}{3} \vec{A}\right) \end{vmatrix} = 0 \]

### Step 6: Solve the Determinant
Expanding the determinant and simplifying will yield a relationship involving \( \lambda \).

### Step 7: Solve for \( \lambda \)
After simplifying the determinant, we will find that:
\[ \lambda = -1 \]

### Final Answer
Thus, the value of \( \lambda \) such that points \( P, Q, R, S \) are coplanar is:
\[ \boxed{-1} \]