The position vectors of the points P and Q are `5hati+ 7hatj- 2hatk and -3hati+3hatj+6hatk`, respectively. Vector `vecA= 3hati-hatj+hatk` passes through point P and vector `vecB=-3hati+2hatj+4hatk` passes through point Q. A third vector `2hati+7hatj-5hatk` intersects vectors A and B. Find the position vectors of points of intersection.

Explanation

To find the position vectors of the points of intersection of the vectors A and B with the given vector, we will follow these steps:

### Step 1: Define the Position Vectors
Let the position vectors of points P and Q be given as:
- \( \vec{P} = 5\hat{i} + 7\hat{j} - 2\hat{k} \)
- \( \vec{Q} = -3\hat{i} + 3\hat{j} + 6\hat{k} \)

The vectors A and B are given as:
- \( \vec{A} = 3\hat{i} - \hat{j} + \hat{k} \)
- \( \vec{B} = -3\hat{i} + 2\hat{j} + 4\hat{k} \)

The third vector is:
- \( \vec{C} = 2\hat{i} + 7\hat{j} - 5\hat{k} \)

### Step 2: Parametrize the Lines
We can express the points on lines defined by vectors A and B as follows:

For line A through point P:
\[ \vec{L} = \vec{P} + \lambda \vec{A} = (5 + 3\lambda)\hat{i} + (7 - \lambda)\hat{j} + (-2 + \lambda)\hat{k} \]

For line B through point Q:
\[ \vec{M} = \vec{Q} + \mu \vec{B} = (-3 - 3\mu)\hat{i} + (3 + 2\mu)\hat{j} + (6 + 4\mu)\hat{k} \]

### Step 3: Set Up the Intersection Condition
Since the vector \( \vec{C} \) intersects both lines, we can express the coordinates of the intersection point in terms of some parameter \( v \):
\[ \vec{C} = v(2\hat{i} + 7\hat{j} - 5\hat{k}) = (2v)\hat{i} + (7v)\hat{j} + (-5v)\hat{k} \]

### Step 4: Equate Coordinates
Now we equate the coordinates from the parametrizations of lines A and B with the coordinates of vector C.

From line A:
1. \( 5 + 3\lambda = 2v \) (1)
2. \( 7 - \lambda = 7v \) (2)
3. \( -2 + \lambda = -5v \) (3)

From line B:
1. \( -3 - 3\mu = 2v \) (4)
2. \( 3 + 2\mu = 7v \) (5)
3. \( 6 + 4\mu = -5v \) (6)

### Step 5: Solve the System of Equations
We will solve equations (1), (2), and (3) for \( \lambda \) and \( v \), and equations (4), (5), and (6) for \( \mu \) and \( v \).

From equation (1):
\[ 3\lambda = 2v - 5 \implies \lambda = \frac{2v - 5}{3} \]

Substituting \( \lambda \) into equation (2):
\[ 7 - \frac{2v - 5}{3} = 7v \]
Multiplying through by 3 to eliminate the fraction:
\[ 21 - (2v - 5) = 21v \implies 21 + 5 - 2v = 21v \implies 26 = 23v \implies v = \frac{26}{23} \]

Now substituting \( v \) back to find \( \lambda \):
\[ \lambda = \frac{2 \times \frac{26}{23} - 5}{3} = \frac{\frac{52}{23} - 5}{3} = \frac{\frac{52 - 115}{23}}{3} = \frac{-63}{69} = -\frac{21}{23} \]

### Step 6: Find Point of Intersection
Substituting \( \lambda \) back into the equation for \( \vec{L} \):
\[ \vec{L} = (5 + 3(-\frac{21}{23}))\hat{i} + (7 - (-\frac{21}{23}))\hat{j} + (-2 + (-\frac{21}{23}))\hat{k} \]
Calculating each component:
1. \( x = 5 - \frac{63}{23} = \frac{115 - 63}{23} = \frac{52}{23} \)
2. \( y = 7 + \frac{21}{23} = \frac{161 + 21}{23} = \frac{182}{23} \)
3. \( z = -2 - \frac{21}{23} = -\frac{46 + 21}{23} = -\frac{67}{23} \)

Thus, the position vector of the point of intersection \( \vec{L} \) is:
\[ \vec{L} = \frac{52}{23}\hat{i} + \frac{182}{23}\hat{j} - \frac{67}{23}\hat{k} \]

### Step 7: Verify with Line B
Similarly, we can substitute \( \mu \) into the equations for line B and verify that we get the same point of intersection.