Find the vector equation of a line passing through `(2, -1, 1)` and parallel to the line whose equation is `(X-3)/(2)=(Y+1)/(7)=(Z-2)/(-3)`.

Answer

To find the vector equation of a line passing through the point \( (2, -1, 1) \) and parallel to the given line, we will follow these steps:

### Step 1: Identify the point and direction ratios
The point through which our line passes is given as \( (2, -1, 1) \). We can represent this point as a position vector \( \mathbf{a} \):
\[ \mathbf{a} = 2\mathbf{i} - 1\mathbf{j} + 1\mathbf{k} \]

The direction ratios of the given line are obtained from the equation:
\[ \frac{X-3}{2} = \frac{Y+1}{7} = \frac{Z-2}{-3} \]
From this, we can identify the direction ratios as \( (2, 7, -3) \). We can represent this as a direction vector \( \mathbf{b} \):
\[ \mathbf{b} = 2\mathbf{i} + 7\mathbf{j} - 3\mathbf{k} \]

### Step 2: Write the vector equation of the line
The vector equation of a line can be expressed as:
\[ \mathbf{r} = \mathbf{a} + \lambda \mathbf{b} \]
where \( \lambda \) is a parameter.

Substituting the values of \( \mathbf{a} \) and \( \mathbf{b} \):
\[ \mathbf{r} = (2\mathbf{i} - 1\mathbf{j} + 1\mathbf{k}) + \lambda (2\mathbf{i} + 7\mathbf{j} - 3\mathbf{k}) \]

### Step 3: Simplify the equation
Now, we can simplify the equation:
\[ \mathbf{r} = 2\mathbf{i} - 1\mathbf{j} + 1\mathbf{k} + \lambda (2\mathbf{i} + 7\mathbf{j} - 3\mathbf{k}) \]
Distributing \( \lambda \):
\[ \mathbf{r} = 2\mathbf{i} - 1\mathbf{j} + 1\mathbf{k} + 2\lambda \mathbf{i} + 7\lambda \mathbf{j} - 3\lambda \mathbf{k} \]
Combining like terms:
\[ \mathbf{r} = (2 + 2\lambda)\mathbf{i} + (-1 + 7\lambda)\mathbf{j} + (1 - 3\lambda)\mathbf{k} \]

### Final Answer
Thus, the vector equation of the line is:
\[ \mathbf{r} = (2 + 2\lambda)\mathbf{i} + (-1 + 7\lambda)\mathbf{j} + (1 - 3\lambda)\mathbf{k} \]