The gradient of the line \(l\) is found by rearranging \(2x + y = k\) to \(y = -2x + k\). Thus, the gradient is \(-2\).

For the curve \(xy = 12\), differentiate implicitly: \(x \frac{dy}{dx} + y = 0\).

At \(P(2, 6)\), substitute \(x = 2\) and \(y = 6\) into the differentiated equation:

\(2 \frac{dy}{dx} + 6 = 0\)

\(\frac{dy}{dx} = -\frac{6}{2} = -3\)

The gradient of the tangent at \(P\) is \(-3\).

The angle \(\theta\) between the line and the tangent is given by:

\(\tan \theta = \left| \frac{-3 - (-2)}{1 + (-3)(-2)} \right| = \left| \frac{-1}{1 + 6} \right| = \frac{1}{7}\)

\(\theta = \tan^{-1} \left( \frac{1}{7} \right) \approx 8.1^\circ\)