Given that $x^2 + y^2 = 14x + 6y + 6$, what is the largest possible value that $3x + 4y$ can have?

$\text{(A)}\ 72 \qquad \text{(B)}\ 73 \qquad \text{(C)}\ 74 \qquad \text{(D)}\ 75\qquad \text{(E)}\ 76$
To solve the problem, we need to find the maximum value of \(3x + 4y\) given the equation of the circle \(x^2 + y^2 = 14x + 6y + 6\).

1. **Rewrite the circle equation in standard form:**
   \[
   x^2 + y^2 = 14x + 6y + 6
   \]
   We complete the square for both \(x\) and \(y\):
   \[
   x^2 - 14x + y^2 - 6y = 6
   \]
   \[
   (x - 7)^2 - 49 + (y - 3)^2 - 9 = 6
   \]
   \[
   (x - 7)^2 + (y - 3)^2 = 64
   \]
   This represents a circle with center \((7, 3)\) and radius \(8\).

2. **Consider the family of lines \(3x + 4y = k\):**
   We want to find the maximum value of \(k\) such that the line is tangent to the circle. The line \(3x + 4y = k\) can be rewritten in slope-intercept form:
   \[
   y = -\frac{3}{4}x + \frac{k}{4}
   \]

3. **Find the distance from the center of the circle to the line:**
   The distance \(d\) from a point \((x_1, y_1)\) to a line \(Ax + By + C = 0\) is given by:
   \[
   d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}
   \]
   For the line \(3x + 4y - k = 0\) and the center \((7, 3)\):
   \[
   d = \frac{|3(7) + 4(3) - k|}{\sqrt{3^2 + 4^2}} = \frac{|21 + 12 - k|}{5} = \frac{|33 - k|}{5}
   \]
   Since the line is tangent to the circle, the distance \(d\) must equal the radius of the circle, which is \(8\):
   \[
   \frac{|33 - k|}{5} = 8
   \]
   \[
   |33 - k| = 40
   \]

4. **Solve for \(k\):**
   \[
   33 - k = 40 \quad \text{or} \quad 33 - k = -40
   \]
   \[
   k = -7 \quad \text{or} \quad k = 73
   \]

5. **Determine the maximum value of \(k\):**
   The maximum value of \(k\) is \(73\).

Conclusion:
\[
\boxed{73}
\]