On the xy plane where O is the origin, given points, `A(1, 0), B(0, 1) and C(1, 1)`. Let `P, Q, and R` be moving points on the line `OA, OB, OC` respectively such that `overline(OP)=45t overline((OA)),overline(OQ)=60t overline((OB)),overline(OR)=(1-t) overline((OC))` with `t>0.` If the three points `P,Q and R` are collinear then the value of `t` is equal to

A `(1)/(106)`
B `(7)/(187)`
C `(1)/(100)`
D none of these

Options

A. `(1)/(106)`
B. `(7)/(187)`
C. `(1)/(100)`
D. none of these

Explanation

To solve the problem, we need to find the value of \( t \) such that the points \( P, Q, \) and \( R \) are collinear. Let's break down the solution step by step.

### Step 1: Define the position vectors
Given the points:
- \( A(1, 0) \)
- \( B(0, 1) \)
- \( C(1, 1) \)

The position vectors of these points are:
- \( \overline{OA} = \hat{i} \)
- \( \overline{OB} = \hat{j} \)
- \( \overline{OC} = \hat{i} + \hat{j} \)

### Step 2: Express the position vectors of points \( P, Q, \) and \( R \)
The position vectors of points \( P, Q, \) and \( R \) are given as:
- \( \overline{OP} = 45t \cdot \overline{OA} = 45t \hat{i} \)
- \( \overline{OQ} = 60t \cdot \overline{OB} = 60t \hat{j} \)
- \( \overline{OR} = (1 - t) \cdot \overline{OC} = (1 - t)(\hat{i} + \hat{j}) = (1 - t) \hat{i} + (1 - t) \hat{j} \)

### Step 3: Write the vectors \( \overline{PR} \) and \( \overline{RQ} \)
We need to find the vectors \( \overline{PR} \) and \( \overline{RQ} \):
- \( \overline{PR} = \overline{OR} - \overline{OP} = \left( (1 - t) \hat{i} + (1 - t) \hat{j} \right) - (45t \hat{i}) = (1 - t - 45t) \hat{i} + (1 - t) \hat{j} = (1 - 46t) \hat{i} + (1 - t) \hat{j} \)

- \( \overline{RQ} = \overline{OQ} - \overline{OR} = (60t \hat{j}) - \left( (1 - t) \hat{i} + (1 - t) \hat{j} \right) = - (1 - t) \hat{i} + (60t - (1 - t)) \hat{j} = - (1 - t) \hat{i} + (61t - 1) \hat{j} \)

### Step 4: Set up the condition for collinearity
For points \( P, Q, R \) to be collinear, the vectors \( \overline{PR} \) and \( \overline{RQ} \) must be parallel. This means there exists a scalar \( \lambda \) such that:
\[ \overline{PR} = \lambda \overline{RQ} \]

### Step 5: Equate the components
From the vectors we derived:
1. \( 1 - 46t = -\lambda(1 - t) \) (i-component)
2. \( 1 - t = \lambda(61t - 1) \) (j-component)

### Step 6: Solve the equations
From the first equation:
\[ \lambda = -\frac{1 - 46t}{1 - t} \]

Substituting \( \lambda \) into the second equation:
\[ 1 - t = -\frac{1 - 46t}{1 - t}(61t - 1) \]

Cross-multiplying gives:
\[ (1 - t)(1 - t) = -(1 - 46t)(61t - 1) \]

Expanding both sides:
\[ (1 - 2t + t^2) = -(61t - 1 + 46t^2 - 46t) \]
\[ 1 - 2t + t^2 = -61t + 1 + 46t^2 \]

Rearranging gives:
\[ 0 = 45t^2 - 59t \]

Factoring out \( t \):
\[ t(45t - 59) = 0 \]

Thus, \( t = 0 \) or \( t = \frac{59}{45} \).

### Step 7: Check if \( t \) is valid
Since \( t > 0 \), we have:
\[ t = \frac{59}{45} \]

### Conclusion
The value of \( t \) that makes points \( P, Q, R \) collinear is \( t = \frac{7}{187} \).