A line with directional cosines prontional to `2, 1,2` meets each of the lines `x = y + a = z and x+ a = 2y = 2z`. The coordinates of each of the points deintersection are given by

Options

A. `(2a, 3a, 3a), (2a, a, a)`
B. `(3a, 2a, 3a), (a, a, a)`
C. `(3a, 2a, 3a), (a, a, 2a)`
D. `(3a, 3a, 3a), (a, a, a)`

Answer

To solve the problem, we need to find the points of intersection of a line with directional cosines proportional to \(2, 1, 2\) with the given lines \(L_1: x = y + a = z\) and \(L_2: x + a = 2y = 2z\).

### Step 1: Understanding the Directional Cosines
The directional cosines of the line are proportional to \(2, 1, 2\). We can represent the line in parametric form as:
\[ x = 2t, \quad y = t, \quad z = 2t \]
where \(t\) is a parameter.

### Step 2: Finding the Parametric Equations for the Given Lines
For the line \(L_1: x = y + a = z\):
- We can express this in parametric form. Let \(y = s\), then:
\[ x = s + a, \quad z = s \]
So, the parametric equations for \(L_1\) are:
\[ x = s + a, \quad y = s, \quad z = s \]

For the line \(L_2: x + a = 2y = 2z\):
- Let \(y = u\), then:
\[ x = 2u - a, \quad z = u \]
So, the parametric equations for \(L_2\) are:
\[ x = 2u - a, \quad y = u, \quad z = u \]

### Step 3: Finding the Intersection with Line \(L_1\)
To find the intersection of our line with \(L_1\), we set the parametric equations equal:
\[ 2t = s + a \quad (1) \]
\[ t = s \quad (2) \]
\[ 2t = s \quad (3) \]

From equation (2), we have \(s = t\). Substituting \(s\) in equations (1) and (3):
From (1):
\[ 2t = t + a \implies t = a \]
From (3):
\[ 2t = t \implies t = 0 \text{ (not possible)} \]

Thus, substituting \(t = a\) back into our parametric equations gives:
\[ x = 2a, \quad y = a, \quad z = 2a \]

### Step 4: Finding the Intersection with Line \(L_2\)
Now, we find the intersection with \(L_2\):
Setting the equations equal:
\[ 2t = 2u - a \quad (4) \]
\[ t = u \quad (5) \]
\[ 2t = u \quad (6) \]

From equation (5), we have \(u = t\). Substituting \(u\) in equations (4) and (6):
From (4):
\[ 2t = 2t - a \implies a = 0 \]
From (6):
\[ 2t = t \implies t = 0 \text{ (not possible)} \]

Thus, substituting \(t = 0\) gives:
\[ x = 0, \quad y = 0, \quad z = 0 \]

### Final Coordinates of Intersection Points
The coordinates of intersection points are:
1. From \(L_1\): \((2a, a, 2a)\)
2. From \(L_2\): \((0, 0, 0)\)

### Summary of the Solution
The coordinates of the points of intersection are:
- For \(L_1\): \((2a, a, 2a)\)
- For \(L_2\): \((0, 0, 0)\)