A line with direction cosines proportional to 2,1,2 meet each of the lines `x=y+a=z and x+a=2y=2z`. The coordinastes of each of the points of intersection are given by
(A) `(3a,2a,3a),(a,a,2a)`
(B) `(3a,2a,3a),(a,a,a)`
(C) `(3a,3a,3a),(a,a,a)`
(D) `(2a,3a,3a),(2a,a,a)`

Answer

To solve the problem, we need to find the points of intersection of a line with direction cosines proportional to 2, 1, 2 with two given lines. Let's break down the solution step by step.

### Step 1: Understand the Given Lines
The two lines given are:
1. \( x = y + a = z \)
2. \( x + a = 2y = 2z \)

### Step 2: Parametrize the Lines
**For Line 1:**
From \( x = y + a = z \), we can express:
- Let \( z = k \) (a parameter), then:
- \( y = k - a \)
- \( x = k \)

So, the parametric equations for line 1 are:
- \( x_1 = k \)
- \( y_1 = k - a \)
- \( z_1 = k \)

**For Line 2:**
From \( x + a = 2y = 2z \), we can express:
- Let \( 2z = \lambda \) (a parameter), then:
- \( z = \frac{\lambda}{2} \)
- \( y = \frac{\lambda}{2} \)
- \( x = \lambda - a \)

So, the parametric equations for line 2 are:
- \( x_2 = \lambda - a \)
- \( y_2 = \frac{\lambda}{2} \)
- \( z_2 = \frac{\lambda}{2} \)

### Step 3: Direction Ratios of the Line
The direction ratios of the line we are considering are proportional to \( 2, 1, 2 \). Thus, we can express the line in terms of a parameter \( t \):
- \( x = 2t \)
- \( y = t \)
- \( z = 2t \)

### Step 4: Setting Up the Equations
Since the points on the line must satisfy the equations of both lines, we equate the coordinates:
1. From the first line:
- \( 2t = k \)
- \( t = k - a \)
- \( 2t = k \)

2. From the second line:
- \( 2t = \lambda - a \)
- \( t = \frac{\lambda}{2} \)
- \( 2t = \frac{\lambda}{2} \)

### Step 5: Solve for \( k \) and \( \lambda \)
From \( 2t = k \):
- \( k = 2t \)

From \( t = k - a \):
- \( t = 2t - a \)
- Rearranging gives \( t = a \)

Substituting \( t = a \) into \( k = 2t \):
- \( k = 2a \)

Now, substituting \( t = a \) into the second line:
- \( 2a = \lambda - a \)
- Rearranging gives \( \lambda = 3a \)

### Step 6: Find the Points of Intersection
**For Line 1:**
Substituting \( k = 2a \):
- \( x_1 = 2a \)
- \( y_1 = 2a - a = a \)
- \( z_1 = 2a \)

So, point \( P \) on line 1 is \( (2a, a, 2a) \).

**For Line 2:**
Substituting \( \lambda = 3a \):
- \( x_2 = 3a - a = 2a \)
- \( y_2 = \frac{3a}{2} \)
- \( z_2 = \frac{3a}{2} \)

So, point \( Q \) on line 2 is \( (2a, \frac{3a}{2}, \frac{3a}{2}) \).

### Conclusion
The coordinates of the points of intersection are:
- Point \( P = (2a, a, 2a) \)
- Point \( Q = (2a, \frac{3a}{2}, \frac{3a}{2}) \)

### Final Answer
The correct option is (D) \( (2a, 3a, 3a), (2a, a, a) \).