Find the cartesian as well as the vector equation of the line passing through :
(i) (-2, 4, -5) and parallel to the line :
` (x + 3)/(3) = (4 -y)/(5) = (z + 8)/(6)`
(ii) (0,-1,4) and parallel to the straight line :
`(-x-2)/(1) = (y + 3)/(7) - (2z - 6)/(3)`.
(iii) (-1, 2,3) and parallel to the line :
`(x - 3)/(2) = (y + 1)/(3) = (z - 1)/(6)`

Answer

To solve the problem, we need to find the Cartesian and vector equations of lines that pass through given points and are parallel to specified lines.

### (i) Line through point (-2, 4, -5) parallel to the line given by:
\[ \frac{x + 3}{3} = \frac{4 - y}{5} = \frac{z + 8}{6} \]

**Step 1: Identify the direction ratios of the given line.**
From the equation, we can extract the direction ratios:
- \( a = 3 \)
- \( b = -5 \)
- \( c = 6 \)

**Step 2: Write the vector equation of the line.**
The vector equation of a line passing through a point \( P(x_1, y_1, z_1) \) with direction ratios \( (a, b, c) \) is given by:
\[ \vec{r} = \vec{r_0} + t \vec{d} \]
Where \( \vec{r_0} \) is the position vector of the point and \( \vec{d} \) is the direction vector.

Here, \( \vec{r_0} = (-2, 4, -5) \) and \( \vec{d} = (3, -5, 6) \).

Thus, the vector equation is:
\[ \vec{r} = (-2, 4, -5) + t(3, -5, 6) \]

**Step 3: Write the Cartesian equation of the line.**
Using the point-direction form:
\[ \frac{x + 2}{3} = \frac{y - 4}{-5} = \frac{z + 5}{6} \]

### (ii) Line through point (0, -1, 4) parallel to the line given by:
\[ \frac{-x - 2}{1} = \frac{y + 3}{7} = \frac{2z - 6}{3} \]

**Step 1: Identify the direction ratios of the given line.**
From the equation, we can extract the direction ratios:
- \( a = -1 \)
- \( b = 7 \)
- \( c = 3 \)

**Step 2: Write the vector equation of the line.**
Using the point \( (0, -1, 4) \):
\[ \vec{r} = (0, -1, 4) + t(-1, 7, 3) \]

**Step 3: Write the Cartesian equation of the line.**
Using the point-direction form:
\[ \frac{x - 0}{-1} = \frac{y + 1}{7} = \frac{z - 4}{3} \]

### (iii) Line through point (-1, 2, 3) parallel to the line given by:
\[ \frac{x - 3}{2} = \frac{y + 1}{3} = \frac{z - 1}{6} \]

**Step 1: Identify the direction ratios of the given line.**
From the equation, we can extract the direction ratios:
- \( a = 2 \)
- \( b = 3 \)
- \( c = 6 \)

**Step 2: Write the vector equation of the line.**
Using the point \( (-1, 2, 3) \):
\[ \vec{r} = (-1, 2, 3) + t(2, 3, 6) \]

**Step 3: Write the Cartesian equation of the line.**
Using the point-direction form:
\[ \frac{x + 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{6} \]

### Summary of Results:
1. For point (-2, 4, -5):
- Vector Equation: \( \vec{r} = (-2, 4, -5) + t(3, -5, 6) \)
- Cartesian Equation: \( \frac{x + 2}{3} = \frac{y - 4}{-5} = \frac{z + 5}{6} \)

2. For point (0, -1, 4):
- Vector Equation: \( \vec{r} = (0, -1, 4) + t(-1, 7, 3) \)
- Cartesian Equation: \( \frac{x - 0}{-1} = \frac{y + 1}{7} = \frac{z - 4}{3} \)

3. For point (-1, 2, 3):
- Vector Equation: \( \vec{r} = (-1, 2, 3) + t(2, 3, 6) \)
- Cartesian Equation: \( \frac{x + 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{6} \)