To find the values of \(m\), equate the line and curve equations: \(mx - 6 = x^2 - 4x + 3\).

Rearrange to form a quadratic: \(x^2 - (4 + m)x + 9 = 0\).

For the line to be tangent, the discriminant must be zero: \(b^2 - 4ac = 0\).

Here, \(a = 1\), \(b = -(4 + m)\), \(c = 9\).

So, \((4 + m)^2 - 4 \times 1 \times 9 = 0\).

\((4 + m)^2 = 36\).

\(4 + m = 6\) or \(4 + m = -6\).

\(m = 2\) or \(m = -10\).

For \(m = 2\), substitute back to find \(x\):

\(x^2 - 6x + 9 = 0\) gives \((x - 3)^2 = 0\), so \(x = 3\).

Substitute \(x = 3\) into \(y = mx - 6\):

\(y = 2 \times 3 - 6 = 0\).

Point is \((3, 0)\).

For \(m = -10\), substitute back to find \(x\):

\(x^2 + 6x + 9 = 0\) gives \((x + 3)^2 = 0\), so \(x = -3\).

Substitute \(x = -3\) into \(y = mx - 6\):

\(y = -10 \times (-3) - 6 = 24\).

Point is \((-3, 24)\).