The position vector of three points are `2veca-vecb+3vecc`, `veca-2vecb+lambdavecc` and `muveca-5vecb` where `veca,vecb,vecc` are non coplanar vectors. The points are collinear when

Options

A. `lambda=-2, mu=(9)/(4)`
B. `lambda=(-9)/(4), mu=2`
C. `lambda=(9)/(4), mu=-2`
D. `lambda=2, mu=-2`

Answer

To determine the values of \( \lambda \) and \( \mu \) for which the points represented by the position vectors \( \vec{p} = 2\vec{a} - \vec{b} + 3\vec{c} \), \( \vec{q} = \vec{a} - 2\vec{b} + \lambda \vec{c} \), and \( \vec{r} = \mu \vec{a} - 5\vec{b} \) are collinear, we follow these steps:

### Step 1: Define the Position Vectors
Let:
- \( \vec{p} = 2\vec{a} - \vec{b} + 3\vec{c} \)
- \( \vec{q} = \vec{a} - 2\vec{b} + \lambda \vec{c} \)
- \( \vec{r} = \mu \vec{a} - 5\vec{b} \)

### Step 2: Establish the Condition for Collinearity
The points are collinear if the vector \( \vec{p} - \vec{q} \) is parallel to the vector \( \vec{q} - \vec{r} \). This can be expressed mathematically as:
\[ \vec{p} - \vec{q} = k(\vec{q} - \vec{r}) \]
for some scalar \( k \).

### Step 3: Calculate \( \vec{p} - \vec{q} \)
\[ \vec{p} - \vec{q} = (2\vec{a} - \vec{b} + 3\vec{c}) - (\vec{a} - 2\vec{b} + \lambda \vec{c}) \]
\[ = (2\vec{a} - \vec{a}) + (-\vec{b} + 2\vec{b}) + (3\vec{c} - \lambda \vec{c}) \]
\[ = \vec{a} + \vec{b} + (3 - \lambda)\vec{c} \]

### Step 4: Calculate \( \vec{q} - \vec{r} \)
\[ \vec{q} - \vec{r} = (\vec{a} - 2\vec{b} + \lambda \vec{c}) - (\mu \vec{a} - 5\vec{b}) \]
\[ = (\vec{a} - \mu \vec{a}) + (-2\vec{b} + 5\vec{b}) + \lambda \vec{c} \]
\[ = (1 - \mu)\vec{a} + 3\vec{b} + \lambda \vec{c} \]

### Step 5: Set Up the Equation
From the collinearity condition:
\[ \vec{a} + \vec{b} + (3 - \lambda)\vec{c} = k((1 - \mu)\vec{a} + 3\vec{b} + \lambda \vec{c}) \]

### Step 6: Compare Coefficients
By comparing coefficients of \( \vec{a} \), \( \vec{b} \), and \( \vec{c} \):

1. Coefficient of \( \vec{a} \):
\[ 1 = k(1 - \mu) \quad \text{(1)} \]

2. Coefficient of \( \vec{b} \):
\[ 1 = 3k \quad \text{(2)} \]

3. Coefficient of \( \vec{c} \):
\[ 3 - \lambda = k\lambda \quad \text{(3)} \]

### Step 7: Solve the Equations
From equation (2):
\[ k = \frac{1}{3} \]

Substituting \( k \) into equation (1):
\[ 1 = \frac{1}{3}(1 - \mu) \implies 1 - \mu = 3 \implies \mu = -2 \]

Substituting \( k \) into equation (3):
\[ 3 - \lambda = \frac{1}{3}\lambda \]
Multiplying through by 3:
\[ 9 - 3\lambda = \lambda \implies 9 = 4\lambda \implies \lambda = \frac{9}{4} \]

### Final Answer
The values of \( \lambda \) and \( \mu \) for which the points are collinear are:
\[ \lambda = \frac{9}{4}, \quad \mu = -2 \]