A straight line is parallel to the lines `3x-y-3=0 and 3x-y+5=0`, and lies between them. Find its equation if its distances from these lines are in the ratio `3 : 5`.

Answer

To solve the problem, we need to find the equation of a straight line that is parallel to the given lines \(3x - y - 3 = 0\) and \(3x - y + 5 = 0\), and lies between them with distances from these lines in the ratio \(3:5\).

### Step-by-Step Solution:

1. **Identify the slopes of the given lines:**
The equations of the lines can be rearranged into slope-intercept form \(y = mx + c\):
- For the line \(3x - y - 3 = 0\):
\[ y = 3x - 3 \]
The slope \(m_1 = 3\).
- For the line \(3x - y + 5 = 0\):
\[ y = 3x + 5 \]
The slope \(m_2 = 3\).

Since both lines have the same slope, any line parallel to them will also have a slope of \(3\).

2. **Set up the equation of the required line:**
The equation of the required line can be expressed as:
\[ y = 3x + c \]
where \(c\) is the y-intercept we need to determine.

3. **Calculate the distances from the required line to the given lines:**
The distance \(d_1\) from the line \(3x - y - 3 = 0\) to the line \(y = 3x + c\) is given by:
\[ d_1 = \frac{|c + 3|}{\sqrt{3^2 + (-1)^2}} = \frac{|c + 3|}{\sqrt{10}} \]
The distance \(d_2\) from the line \(3x - y + 5 = 0\) to the line \(y = 3x + c\) is:
\[ d_2 = \frac{|c - 5|}{\sqrt{10}} \]

4. **Set up the ratio of the distances:**
We know that the ratio of the distances \(d_1\) to \(d_2\) is \(3:5\):
\[ \frac{d_1}{d_2} = \frac{3}{5} \]
Substituting the expressions for \(d_1\) and \(d_2\):
\[ \frac{\frac{|c + 3|}{\sqrt{10}}}{\frac{|c - 5|}{\sqrt{10}}} = \frac{3}{5} \]
This simplifies to:
\[ \frac{|c + 3|}{|c - 5|} = \frac{3}{5} \]

5. **Cross-multiply to eliminate the fraction:**
\[ 5|c + 3| = 3|c - 5| \]

6. **Consider the cases for the absolute values:**
We need to consider two cases for the absolute values.

**Case 1:** \(c + 3 \geq 0\) and \(c - 5 \geq 0\) (i.e., \(c \geq 5\)):
\[ 5(c + 3) = 3(c - 5) \]
Simplifying gives:
\[ 5c + 15 = 3c - 15 \implies 2c = -30 \implies c = -15 \quad \text{(not valid since \(c < 5\))} \]

**Case 2:** \(c + 3 \geq 0\) and \(c - 5 < 0\) (i.e., \(-3 \leq c < 5\)):
\[ 5(c + 3) = -3(c - 5) \]
Simplifying gives:
\[ 5c + 15 = -3c + 15 \implies 8c = 0 \implies c = 0 \]

**Case 3:** \(c + 3 < 0\) and \(c - 5 < 0\) (i.e., \(c < -3\)):
\[ -5(c + 3) = -3(c - 5) \]
Simplifying gives:
\[ -5c - 15 = -3c + 15 \implies -2c = 30 \implies c = -15 \]

7. **Final equations of the required lines:**
We found two possible values for \(c\): \(0\) and \(-15\). Thus, the equations of the required lines are:
- For \(c = 0\): \(y = 3x + 0 \implies y = 3x\)
- For \(c = -15\): \(y = 3x - 15\)

### Final Answer:
The required equations of the lines are:
1. \(y = 3x\)
2. \(y = 3x - 15\)