A line with direction cosines proportional to 2,1,2 meet each of the lines `x=y+a=z nd x+a=2y=2z`. The coordinastes of each of the points of intersection are given by

Options

A. `(3a, 3a, 3a) , (a,a,a)`
B. `(3a,2a, 3a), (a,a,a)`
C. `(3a,2a,3a),(a,a,2a)`
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 first line is given by the equations:
\[ x = y + a = z \]
This can be rewritten as:
\[ x = y + a \quad \text{and} \quad z = y + a \]

The second line is given by:
\[ x + a = 2y = 2z \]
This can be rewritten as:
\[ x + a = 2y \quad \text{and} \quad 2y = 2z \]
From \(2y = 2z\), we can simplify to:
\[ y = z \]
Thus, we can express the second line as:
\[ x + a = 2y \quad \text{and} \quad z = y \]

### Step 2: Parameterize the Lines
For the first line, we can use a parameter \(t\):
\[ x = t, \quad y = t - a, \quad z = t \]

For the second line, we can use a parameter \(s\):
\[ x = 2s - a, \quad y = s, \quad z = s \]

### Step 3: Find the Direction Ratios
The direction cosines of the line we are interested in are proportional to \(2, 1, 2\). Thus, we can express the direction ratios as:
\[ L = 2k, \quad M = k, \quad N = 2k \]
for some constant \(k\).

### Step 4: Set Up the Intersection Conditions
To find the points of intersection, we need to equate the coordinates from both parameterizations:

1. From the first line:
\[ t = 2s - a \quad (1) \]
\[ t - a = s \quad (2) \]
\[ t = s \quad (3) \]

### Step 5: Solve the System of Equations
From equation (3), we have \(t = s\). Substituting \(s\) into equation (2):
\[ t - a = t \implies -a = 0 \implies a = 0 \]

Now substituting \(a = 0\) back into equation (1):
\[ t = 2s \implies t = 2t \implies t = 0 \implies s = 0 \]

### Step 6: Find the Coordinates of Intersection
Substituting \(t = 0\) into the parameterization of the first line:
\[ x = 0, \quad y = 0 - a = 0, \quad z = 0 \]
Thus, the point of intersection for the first line is:
\[ (0, 0, 0) \]

Substituting \(s = 0\) into the parameterization of the second line:
\[ x = 2(0) - a = -a, \quad y = 0, \quad z = 0 \]
Thus, the point of intersection for the second line is:
\[ (-a, 0, 0) \]

### Step 7: Conclusion
The coordinates of the points of intersection are:
1. From the first line: \((0, 0, 0)\)
2. From the second line: \((-a, 0, 0)\)

### Final Answer
The coordinates of the points of intersection are \((0, 0, 0)\) and \((-a, 0, 0)\).

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