(A ) Find the vector and cartesian equations of the line through the point (5,2,-4) and which is parallel to the vector 3 `hat(i) + 2 hat(j) - 8 hat(k)`.
(b) Find the equation of a line passing through the point P(2, -1, 3) and perpendicular to the lines :
`vec(r) = (hat(i) + hat(j) - hat(k) ) + lambda (2 hat(i) - 2 hat(j) + hat(k))`
and `vec(r) = (2 hat(i) - hat(j) - 3 hat(k) ) + mu (hat(i) + 2 hat(j) + 2 hat(k))`.

Answer

To solve the given problem, we will break it down into two parts as specified in the question.

### Part (A)

We need to find the vector and Cartesian equations of the line that passes through the point \( (5, 2, -4) \) and is parallel to the vector \( \vec{b} = 3\hat{i} + 2\hat{j} - 8\hat{k} \).

**Step 1: Write the vector equation of the line**

The vector equation of a line can be expressed as:
\[ \vec{r} = \vec{a} + t\vec{b} \]
where \( \vec{a} \) is a position vector of a point on the line, \( \vec{b} \) is the direction vector, and \( t \) is a scalar parameter.

Here, \( \vec{a} = 5\hat{i} + 2\hat{j} - 4\hat{k} \) and \( \vec{b} = 3\hat{i} + 2\hat{j} - 8\hat{k} \).

Thus, the vector equation becomes:
\[ \vec{r} = (5\hat{i} + 2\hat{j} - 4\hat{k}) + t(3\hat{i} + 2\hat{j} - 8\hat{k}) \]

**Step 2: Expand the vector equation**

Expanding the equation gives:
\[ \vec{r} = (5 + 3t)\hat{i} + (2 + 2t)\hat{j} + (-4 - 8t)\hat{k} \]

**Step 3: Write the Cartesian form**

The Cartesian form of the line can be derived from the vector equation. The general form is:
\[ \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} \]
where \( (x_1, y_1, z_1) \) is the point on the line and \( (a, b, c) \) are the direction ratios.

Here, we have:
- Point: \( (5, 2, -4) \)
- Direction ratios: \( (3, 2, -8) \)

Thus, the Cartesian equation becomes:
\[ \frac{x - 5}{3} = \frac{y - 2}{2} = \frac{z + 4}{-8} \]

### Part (B)

We need to find the equation of a line passing through the point \( P(2, -1, 3) \) and perpendicular to the given lines.

**Step 1: Identify direction vectors of the given lines**

The first line is given by:
\[ \vec{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(2\hat{i} - 2\hat{j} + \hat{k}) \]
The direction vector \( \vec{b_1} = 2\hat{i} - 2\hat{j} + \hat{k} \).

The second line is given by:
\[ \vec{r} = (2\hat{i} - \hat{j} - 3\hat{k}) + \mu(\hat{i} + 2\hat{j} + 2\hat{k}) \]
The direction vector \( \vec{b_2} = \hat{i} + 2\hat{j} + 2\hat{k} \).

**Step 2: Find the direction vector of the required line**

Let the direction vector of the required line be \( \vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k} \).

Since the line is perpendicular to both \( \vec{b_1} \) and \( \vec{b_2} \), we have the following conditions:
1. \( \vec{b} \cdot \vec{b_1} = 0 \)
2. \( \vec{b} \cdot \vec{b_2} = 0 \)

This gives us the equations:
1. \( 2b_1 - 2b_2 + b_3 = 0 \) (Equation 1)
2. \( b_1 + 2b_2 + 2b_3 = 0 \) (Equation 2)

**Step 3: Solve the equations**

From Equation 1, we can express \( b_3 \) in terms of \( b_1 \) and \( b_2 \):
\[ b_3 = -2b_1 + 2b_2 \]

Substituting \( b_3 \) into Equation 2:
\[ b_1 + 2b_2 + 2(-2b_1 + 2b_2) = 0 \]
\[ b_1 + 2b_2 - 4b_1 + 4b_2 = 0 \]
\[ -3b_1 + 6b_2 = 0 \implies b_1 = 2b_2 \]

Let \( b_2 = k \), then \( b_1 = 2k \) and substituting into the expression for \( b_3 \):
\[ b_3 = -2(2k) + 2k = -4k + 2k = -2k \]

Thus, the direction vector \( \vec{b} \) can be expressed as:
\[ \vec{b} = 2k\hat{i} + k\hat{j} - 2k\hat{k} = k(2\hat{i} + \hat{j} - 2\hat{k}) \]

**Step 4: Write the vector equation of the line**

The vector equation of the line passing through point \( P(2, -1, 3) \) is:
\[ \vec{r} = (2\hat{i} - \hat{j} + 3\hat{k}) + \lambda(2\hat{i} + \hat{j} - 2\hat{k}) \]

### Final Answers

**Part (A):**
- Vector equation: \( \vec{r} = (5\hat{i} + 2\hat{j} - 4\hat{k}) + t(3\hat{i} + 2\hat{j} - 8\hat{k}) \)
- Cartesian equation: \( \frac{x - 5}{3} = \frac{y - 2}{2} = \frac{z + 4}{-8} \)

**Part (B):**
- Vector equation: \( \vec{r} = (2\hat{i} - \hat{j} + 3\hat{k}) + \lambda(2\hat{i} + \hat{j} - 2\hat{k}) \)