To find the values of \(k\) for which the line is tangent to the curve, we set the equations equal to each other:

\(kx + 6 = x^2 + 3x + 2k\)

Rearrange to form a quadratic equation:

\(x^2 + (3-k)x + (2k-6) = 0\)

For the line to be tangent to the curve, the discriminant of this quadratic must be zero:

\(b^2 - 4ac = 0\)

Here, \(a = 1\), \(b = 3-k\), \(c = 2k-6\).

Substitute into the discriminant formula:

\((3-k)^2 - 4(1)(2k-6) = 0\)

Simplify:

\((3-k)^2 - 8k + 24 = 0\)

\(9 - 6k + k^2 - 8k + 24 = 0\)

\(k^2 - 14k + 33 = 0\)

Factorize the quadratic:

\((3-k)(11-k) = 0\)

Thus, \(k = 3\) or \(k = 11\).