To find the perpendicular form of the given straight line, we first need to understand the equation of the line and determine its slope. The given line is:

\(\sqrt{3}x + 2y = 7\)

Let's convert this equation to the slope-intercept form, \(y = mx + c\), where \(m\) is the slope.

1. Rearrange the equation: \(2y = -\sqrt{3}x + 7\)
2. Divide all terms by 2 to solve for \(y\): \(y = -\frac{\sqrt{3}}{2}x + \frac{7}{2}\)
3. From this form, the slope \(m\) is: \(-\frac{\sqrt{3}}{2}\)

For a perpendicular line, the slope will be the negative reciprocal of the original slope. If the slope of the original line is \(-\frac{\sqrt{3}}{2}\), the perpendicular slope \(m'\) will be:

\(m' = \frac{2}{\sqrt{3}}\)

Next, we need to express the equation of the line with this new slope in the required perpendicular form:

1. Multiplying the line equation by a scalar that makes the coefficients of \(x\) and \(y\) unit vectors: \(\frac{\sqrt{3}}{\sqrt{7}}x + \frac{2}{\sqrt{7}}y = k\)
2. Solving for \(k\) using appropriate normalization, we set: \(k = \sqrt{7}\)

Thus, the correct equation in the perpendicular form to the given line is:

\(\frac{\sqrt{3}}{\sqrt{7}}x + \frac{2}{\sqrt{7}}y = \sqrt{7}\)

This matches with the option provided:

• \(\frac{\sqrt{3}}{\sqrt{7}}x + \frac{2}{\sqrt{7}}y = \sqrt{7}\)

Hence, this option is the perpendicular form of the given line equation.