(A) The cartesian equations of a line are :
`(i) (x - 5)/(3) = (y + 4)/(7) = (z - 6)/(2) `
(ii) `(x + 3)/(2) = (y - 5)/(4) = (z + 6)/(2)` . Find the vector equations of the lines.
(b) find the vector equation of the line passing through the point A (1,2, - 1) and parallel to the line :
5 x - 25 = 14 - 7y = 35 z.

Answer

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

### Part (A)

#### Step 1: Convert the first Cartesian equation to vector form
The first line is given by the equation:
\[ \frac{x - 5}{3} = \frac{y + 4}{7} = \frac{z - 6}{2} \]
This can be rewritten in the standard form:
\[ \frac{x - 5}{3} = \frac{y + 4}{7} = \frac{z - 6}{2} = t \]
From this, we can express \(x\), \(y\), and \(z\) in terms of \(t\):
\[ x = 5 + 3t, \quad y = -4 + 7t, \quad z = 6 + 2t \]
The position vector \(A\) (point through which the line passes) is:
\[ A = 5\hat{i} - 4\hat{j} + 6\hat{k} \]
The direction vector \(B\) is:
\[ B = 3\hat{i} + 7\hat{j} + 2\hat{k} \]
Thus, the vector equation of the line is:
\[ \vec{r} = \vec{A} + \lambda \vec{B} \]
Substituting the values, we get:
\[ \vec{r} = (5\hat{i} - 4\hat{j} + 6\hat{k}) + \lambda(3\hat{i} + 7\hat{j} + 2\hat{k}) \]

#### Step 2: Convert the second Cartesian equation to vector form
The second line is given by the equation:
\[ \frac{x + 3}{2} = \frac{y - 5}{4} = \frac{z + 6}{2} \]
This can also be rewritten in the standard form:
\[ \frac{x + 3}{2} = \frac{y - 5}{4} = \frac{z + 6}{2} = s \]
From this, we can express \(x\), \(y\), and \(z\) in terms of \(s\):
\[ x = -3 + 2s, \quad y = 5 + 4s, \quad z = -6 + 2s \]
The position vector \(A\) is:
\[ A = -3\hat{i} + 5\hat{j} - 6\hat{k} \]
The direction vector \(B\) is:
\[ B = 2\hat{i} + 4\hat{j} + 2\hat{k} \]
Thus, the vector equation of the second line is:
\[ \vec{r} = \vec{A} + \lambda \vec{B} \]
Substituting the values, we get:
\[ \vec{r} = (-3\hat{i} + 5\hat{j} - 6\hat{k}) + \lambda(2\hat{i} + 4\hat{j} + 2\hat{k}) \]

### Part (B)

#### Step 3: Find the vector equation of the line passing through point A and parallel to the given line
The line is given by:
\[ 5x - 25 = 14 - 7y = 35z \]
We can rewrite it in standard form:
\[ \frac{x - 5}{7} = \frac{y - 2}{-5} = \frac{z - 0}{1} = k \]
From this, we can express \(x\), \(y\), and \(z\) in terms of \(k\):
\[ x = 5 + 7k, \quad y = 2 - 5k, \quad z = k \]
The direction vector \(B\) is:
\[ B = 7\hat{i} - 5\hat{j} + \hat{k} \]
The point through which the new line passes is \(A(1, 2, -1)\), which gives us the position vector:
\[ \vec{A} = 1\hat{i} + 2\hat{j} - 1\hat{k} \]
Thus, the vector equation of the required line is:
\[ \vec{r} = \vec{A} + \lambda \vec{B} \]
Substituting the values, we get:
\[ \vec{r} = (1\hat{i} + 2\hat{j} - 1\hat{k}) + \lambda(7\hat{i} - 5\hat{j} + \hat{k}) \]

### Final Answers
1. For the first line:
\[ \vec{r} = (5\hat{i} - 4\hat{j} + 6\hat{k}) + \lambda(3\hat{i} + 7\hat{j} + 2\hat{k}) \]
2. For the second line:
\[ \vec{r} = (-3\hat{i} + 5\hat{j} - 6\hat{k}) + \lambda(2\hat{i} + 4\hat{j} + 2\hat{k}) \]
3. For the line through point \(A(1, 2, -1)\) and parallel to the given line:
\[ \vec{r} = (1\hat{i} + 2\hat{j} - 1\hat{k}) + \lambda(7\hat{i} - 5\hat{j} + \hat{k}) \]