1. **State the problem:** We need to find the equation of the line $k$ that passes through the point $P(6, -3)$ and is perpendicular to the line $l$ given by the equation $$3x - 4y = 5.$$\n\n2. **Find the slope of line $l$:** Rewrite $l$ in slope-intercept form $y = mx + c$.\n$$3x - 4y = 5 \implies -4y = -3x + 5 \implies y = \frac{3}{4}x - \frac{5}{4}.$$\nSo, the slope of $l$ is $m_l = \frac{3}{4}$.\n\n3. **Find the slope of line $k$:** Lines perpendicular to each other have slopes that are negative reciprocals.\n$$m_k = -\frac{1}{m_l} = -\frac{1}{\frac{3}{4}} = -\frac{4}{3}.$$\n\n4. **Use point-slope form to find equation of $k$:** The point-slope form is\n$$y - y_1 = m(x - x_1),$$\nwhere $(x_1, y_1) = (6, -3)$ and $m = -\frac{4}{3}$.\n$$y - (-3) = -\frac{4}{3}(x - 6) \implies y + 3 = -\frac{4}{3}x + 8.$$\n\n5. **Rewrite in standard form $ax + by + c = 0$ with integers:**\nMultiply both sides by 3 to clear the fraction:\n$$3(y + 3) = 3\left(-\frac{4}{3}x + 8\right) \implies 3y + 9 = -4x + 24.$$\n\nBring all terms to one side:\n$$4x + 3y + 9 - 24 = 0 \implies 4x + 3y - 15 = 0.$$\n\n6. **Final answer:** The equation of line $k$ is\n$$\boxed{4x + 3y - 15 = 0}.$$