A line has y-intercept  $(0,3)$ and forms a right angle to the line $2x + y  =  3$. Find the x-intercept of the line.

$\textbf{(A) }(4,0)\quad\textbf{(B) }(6,0)\quad\textbf{(C) }(-4,0)\quad\textbf{(D) }(-6,0)\quad\textbf{(E) }\text{none of the above}$
1. **Identify the slope of the given line:**
   The given line is \(2x + y = 3\). To find its slope, we rewrite it in the slope-intercept form \(y = mx + b\):
   \[
   y = -2x + 3
   \]
   Here, the slope \(m\) is \(-2\).

2. **Determine the slope of the perpendicular line:**
   Perpendicular lines have slopes that are negative reciprocals of each other. Therefore, the slope of the line perpendicular to \(y = -2x + 3\) is:
   \[
   m_{\text{perpendicular}} = -\frac{1}{-2} = \frac{1}{2}
   \]

3. **Form the equation of the perpendicular line:**
   The perpendicular line has a slope of \(\frac{1}{2}\) and a y-intercept of 3 (given as \((0, 3)\)). Using the slope-intercept form \(y = mx + b\), we get:
   \[
   y = \frac{1}{2}x + 3
   \]

4. **Find the x-intercept of the perpendicular line:**
   The x-intercept occurs where \(y = 0\). Substitute \(y = 0\) into the equation \(y = \frac{1}{2}x + 3\) and solve for \(x\):
   \[
   0 = \frac{1}{2}x + 3
   \]
   \[
   \frac{1}{2}x = -3
   \]
   \[
   x = -3 \times 2
   \]
   \[
   x = -6
   \]

Therefore, the x-intercept of the line is \((-6, 0)\).

The final answer is \(\boxed{(-6, 0)}\)