To find the values of \(c\) and coordinates of \(P\), we set the equations equal since the line is tangent to the curve:

\(cx^2 + 2x - 3 = 6x - c\)

Rearrange to form a quadratic equation:

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

\(cx^2 - 4x + (c - 3) = 0\)

For the line to be tangent, the discriminant must be zero:

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

Here, \(a = c\), \(b = -4\), \(c = c - 3\).

\((-4)^2 - 4(c)(c - 3) = 0\)

\(16 - 4c^2 + 12c = 0\)

\(4c^2 - 12c - 16 = 0\)

Divide by 4:

\(c^2 - 3c - 4 = 0\)

Factorize:

\((c - 4)(c + 1) = 0\)

Thus, \(c = 4\) or \(c = -1\).

For \(c = 4\):

Substitute \(c = 4\) into the quadratic equation:

\(4x^2 - 4x + 1 = 0\)

\((2x - 1)^2 = 0\)

\(x = \frac{1}{2}\)

Substitute \(x = \frac{1}{2}\) into \(y = 6x - c\):

\(y = 6 \times \frac{1}{2} - 4 = -1\)

Coordinates of \(P\) are \(\left( \frac{1}{2}, -1 \right)\).

For \(c = -1\):

Substitute \(c = -1\) into the quadratic equation:

\(x^2 + 4x + 4 = 0\)

\((x + 2)^2 = 0\)

\(x = -2\)

Substitute \(x = -2\) into \(y = 6x - c\):

\(y = 6 \times (-2) + 1 = -11\)

Coordinates of \(P\) are \((-2, -11)\).