The equations of sides $$AB$$ and $$AC$$ of a triangle $$ABC$$ are $$(\lambda + 1)x + \lambda y = 4$$ and $$\lambda x + (1 - \lambda)y + \lambda = 0$$ respectively. Its vertex $$A$$ is on the $$y$$-axis and its orthocentre is $$(1, 2)$$. The length of the tangent from the point $$C$$ to the part of the parabola $$y^2 = 6x$$ in the first quadrant is

Line $$AB$$: $$(\lambda+1)x + \lambda y = 4$$

Line $$AC$$: $$\lambda x + (1-\lambda)y + \lambda = 0$$

Since vertex $$A$$ lies on the y-axis, we set $$x = 0$$: 

From $$AB$$: $$\lambda y = 4 \implies y = \frac{4}{\lambda}$$

From $$AC$$: $$(1-\lambda)y + \lambda = 0 \implies y = \frac{\lambda}{\lambda - 1}$$

$$\frac{4}{\lambda} = \frac{\lambda}{\lambda - 1}$$

$$4\lambda - 4 = \lambda^2 \implies \lambda^2 - 4\lambda + 4 = 0 \implies (\lambda - 2)^2 = 0 \implies \lambda = 2$$

$$A = (0, 2)$$

Equation of $$AB$$: $$3x + 2y = 4 \implies \text{Slope } m_{AB} = -\frac{3}{2}$$

Equation of $$AC$$: $$2x - y + 2 = 0 \implies \text{Slope } m_{AC} = 2$$

Let the orthocenter be $$H(1, 2)$$.

The altitude from vertex $$B$$ passes through $$H(1, 2)$$ and is perpendicular to side $$AC$$.

Since the slope of $$AC$$ is $$2$$, the slope of the altitude $$BH$$ is $$-\frac{1}{2}$$.

Equation of altitude $$BH$$: $$y - 2 = -\frac{1}{2}(x - 1) \implies x + 2y = 5$$

Vertex $$B$$ is the intersection of side $$AB$$ ($$3x + 2y = 4$$) and altitude $$BH$$ ($$x + 2y = 5$$): $$\implies B = \left(-\frac{1}{2}, \frac{11}{4}\right)$$

Since $$m_{AB} = -\frac{3}{2}$$, the slope of the altitude $$CH$$ is $$\frac{2}{3}$$. Equation of altitude $$CH$$: $$y - 2 = \frac{2}{3}(x - 1) \implies 2x - 3y + 4 = 0$$

Vertex $$C$$ is the intersection of side $$AC$$ ($$2x - y + 2 = 0$$) and altitude $$CH$$ ($$2x - 3y + 4 = 0$$): $$C = \left(-\frac{1}{2}, 1\right)$$

The parabola is $$y^2 = 6x \implies 4a = 6 \implies a = \frac{3}{2}$$

$$y = mx + \frac{a}{m} \implies y = mx + \frac{3}{2m}$$

$$1 = m\left(-\frac{1}{2}\right) + \frac{3}{2m}$$ (Passes through $$C$$)

$$1 = -\frac{m}{2} + \frac{3}{2m}$$

$$2m = -m^2 + 3 \implies m^2 + 2m - 3 = 0$$

$$(m + 3)(m - 1) = 0 \implies m = 1 \quad \text{or} \quad m = -3$$

The point of contact formula is $$\left(\frac{a}{m^2}, \frac{2a}{m}\right) = \left(\frac{3}{2m^2}, \frac{3}{m}\right)$$. For the point to lie in the first quadrant, the y-coordinate must be positive, which means $$m > 0$$. Thus, $$m = 1$$.

$$P = \left(\frac{3}{2(1)^2}, \frac{3}{1}\right) = \left(\frac{3}{2}, 3\right)$$

Length of $$CP$$: $$\text{Length} = \sqrt{\left(\frac{3}{2} - \left(-\frac{1}{2}\right)\right)^2 + (3 - 1)^2}$$

$$\text{Length} = \sqrt{(2)^2 + (2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$$