The problem asks for the slope of the line that is perpendicular to the line $3x + 12y = 6$.

1. Problem Description

The problem asks for the slope of the line that is perpendicular to the line $3x + 12y = 6$.

2. Solution Steps

First, we need to find the slope of the given line. To do this, we can rewrite the equation in slope-intercept form, which is $y = mx + b$, where $m$ is the slope and $b$ is the y-intercept.

Starting with the equation $3x + 12y = 6$, we can isolate $y$:

$12y = -3x + 6$

$y = \frac{-3x + 6}{12}$

$y = -\frac{3}{12}x + \frac{6}{12}$

$y = -\frac{1}{4}x + \frac{1}{2}$

So the slope of the given line is $m_1 = -\frac{1}{4}$.

The slope of a line perpendicular to another line is the negative reciprocal of the slope of the original line. If $m_1$ is the slope of the original line and $m_2$ is the slope of the perpendicular line, then $m_2 = -\frac{1}{m_1}$.

In this case, $m_1 = -\frac{1}{4}$, so the slope of the perpendicular line is:

$m_2 = -\frac{1}{-\frac{1}{4}} = -(-4) = 4$.

3. Final Answer

The slope of the line perpendicular to the line $3x + 12y = 6$ is

4. Therefore, the answer is (c).