The line joining the points `6veca-4vecb+4vecc, -4vecc` and the line joining the points `-veca-2vecb-3vecc, veca+2vecb-5vecc`intersect at

Options

A. `-4veca`
B. `4veca-vecb-vecc`
C. `4vecc`
D. none of these

Explanation

To find the intersection of the two lines given in the question, we will follow these steps:

### Step 1: Identify the Points
The points for the first line are given as:
- Point A: \(6\vec{a} - 4\vec{b} + 4\vec{c}\)
- Point B: \(-4\vec{c}\)

The points for the second line are given as:
- Point C: \(-\vec{a} - 2\vec{b} - 3\vec{c}\)
- Point D: \(\vec{a} + 2\vec{b} - 5\vec{c}\)

### Step 2: Write the Parametric Equations of the Lines
For Line 1 (joining points A and B):
- The direction vector can be calculated as:
\[ \text{Direction vector} = B - A = (-4\vec{c}) - (6\vec{a} - 4\vec{b} + 4\vec{c}) = -6\vec{a} + 4\vec{b} - 8\vec{c} \]
- The parametric equations of Line 1 can be written as:
\[ \frac{x - 6}{6} = \frac{y + 4}{-4} = \frac{z - 4}{-8} = k \]
From this, we can derive:
\[ x = 6k + 6, \quad y = -4k - 4, \quad z = -8k + 4 \]

For Line 2 (joining points C and D):
- The direction vector can be calculated as:
\[ \text{Direction vector} = D - C = (\vec{a} + 2\vec{b} - 5\vec{c}) - (-\vec{a} - 2\vec{b} - 3\vec{c}) = 2\vec{a} + 4\vec{b} - 2\vec{c} \]
- The parametric equations of Line 2 can be written as:
\[ \frac{x + 1}{-1} = \frac{y + 2}{-2} = \frac{z + 3}{-3} = r \]
From this, we can derive:
\[ x = -r - 1, \quad y = -2r - 2, \quad z = -3r - 3 \]

### Step 3: Set the Equations Equal to Each Other
Since the lines intersect, we can set the equations equal to each other:
1. From the x-coordinates:
\[ 6k + 6 = -r - 1 \quad \text{(1)} \]
2. From the y-coordinates:
\[ -4k - 4 = -2r - 2 \quad \text{(2)} \]
3. From the z-coordinates:
\[ -8k + 4 = -3r - 3 \quad \text{(3)} \]

### Step 4: Solve the System of Equations
From equation (1):
\[ 6k + r = -7 \quad \text{(4)} \]

From equation (2):
\[ -4k + 2r = 2 \quad \Rightarrow \quad 2r = 4k + 2 \quad \Rightarrow \quad r = 2k + 1 \quad \text{(5)} \]

Substituting equation (5) into equation (4):
\[ 6k + (2k + 1) = -7 \\ 8k + 1 = -7 \\ 8k = -8 \\ k = -1 \]

Now substituting \(k = -1\) into equation (5):
\[ r = 2(-1) + 1 = -2 + 1 = -1 \]

### Step 5: Find the Intersection Point
Now we can find the coordinates of the intersection point by substituting \(k = -1\) into the parametric equations of Line 1:
\[ x = 6(-1) + 6 = 0 \\ y = -4(-1) - 4 = 0 \\ z = -8(-1) + 4 = 4 + 4 = 0 \]

Thus, the intersection point is:
\[ (0, 0, 0) \quad \text{or} \quad 0\vec{c} \]

### Final Answer
The lines intersect at the point \(0\vec{c}\).

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