This problem is a comprehensive test of coordinate geometry concepts, requiring us to first determine the geometric properties of a triangle (vertices, orthocenter) and then apply these to a conic section (parabola) to find the length of a tangent. We will systematically break down the solution into logical steps.

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1. Determining the Value of $\lambda$ and Coordinates of Vertex A

• Key Concept: A point lying on the y-axis has an x-coordinate of 0. The intersection point of two lines satisfies both their equations.
• Explanation: Vertex A is the common point of the lines AB and AC. We are given that A lies on the y-axis, meaning its x-coordinate is 0. We can substitute $x=0$ into the equations of both lines to find the y-coordinate of A in terms of $\lambda$. Since A is a unique point, these y-coordinates must be equal, allowing us to solve for $\lambda$.

Let the coordinates of vertex A be $(0, y_A)$. Substitute $x=0$ into the equation of side AB: $(\lambda+1)(0) + \lambda y_A = 4$ $\lambda y_A = 4 \quad \Rightarrow \quad y_A = \frac{4}{\lambda} \quad (\text{Equation 1})$

• Tip: Note that $\lambda \neq 0$ for $y_A$ to be defined. If $\lambda=0$, the equation of AB becomes $x=4$, which is a vertical line. If A is on the y-axis ($x=0$), then $x=4$ and $x=0$ cannot intersect, so $\lambda \neq 0$.

Now, substitute $x=0$ into the equation of side AC: $\lambda(0) + (1-\lambda)y_A + \lambda = 0$ $(1-\lambda)y_A = -\lambda \quad \Rightarrow \quad y_A = \frac{-\lambda}{1-\lambda} \quad (\text{Equation 2})$

• Tip: Note that $\lambda \neq 1$ for $y_A$ to be defined. If $\lambda=1$, the equation of AC becomes $x+1=0$, or $x=-1$, which is a vertical line. If A is on the y-axis ($x=0$), then $x=-1$ and $x=0$ cannot intersect, so $\lambda \neq 1$.

Equating the two expressions for $y_A$ (from Equation 1 and Equation 2): $\frac{4}{\lambda} = \frac{-\lambda}{1-\lambda}$ $4(1-\lambda) = -\lambda^2$ $4 - 4\lambda = -\lambda^2$ $\lambda^2 - 4\lambda + 4 = 0$ This is a perfect square quadratic equation: $(\lambda-2)^2 = 0$ Therefore, $\lambda = 2$.

Now, substitute $\lambda=2$ back into Equation 1 (or Equation 2) to find $y_A$: $y_A = \frac{4}{2} = 2$ So, the coordinates of vertex A are $(0, 2)$.

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2. Equations of Sides AB and AC

• Key Concept: Substitute the determined value of $\lambda$ into the given general equations to find the specific equations of the lines.

Substitute $\lambda=2$ into the equation of side AB: $(\lambda+1)x+\lambda y=4 \quad \Rightarrow \quad (2+1)x + 2y = 4$ $3x + 2y = 4 \quad (\text{Equation of AB})$

Substitute $\lambda=2$ into the equation of side AC: $\lambda x+(1-\lambda)y+\lambda=0 \quad \Rightarrow \quad 2x + (1-2)y + 2 = 0$ $2x - y + 2 = 0 \quad (\text{Equation of AC})$

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3. Determining the Coordinates of Vertex C

• Key Concept: The orthocentre H is the intersection of the altitudes of a triangle. An altitude from a vertex is perpendicular to the opposite side.
• Explanation: We know the orthocentre H is $(1, 2)$. The altitude from vertex C, denoted as CH, is perpendicular to the side AB. We can find the slope of AB, then the slope of CH, and use the point-slope form to write the equation of CH. Vertex C will be the intersection of side AC and the altitude CH.

First, find the slope of side AB. From $3x+2y=4$, rearrange to $y = -\frac{3}{2}x + 2$. The slope of AB, $m_{AB} = -\frac{3}{2}$.

Since the altitude CH is perpendicular to AB, the product of their slopes is -1: $m_{CH} \cdot m_{AB} = -1 \quad \Rightarrow \quad m_{CH} \cdot \left(-\frac{3}{2}\right) = -1$ $m_{CH} = \frac{2}{3}$

Now, write the equation of the altitude CH using the point-slope form $y-y_1 = m(x-x_1)$, with H$(1, 2)$ as $(x_1, y_1)$ and $m_{CH} = \frac{2}{3}$: $y - 2 = \frac{2}{3}(x - 1)$ $3(y - 2) = 2(x - 1)$ $3y - 6 = 2x - 2$ $2x - 3y + 4 = 0 \quad (\text{Equation of CH})$

Vertex C is the intersection of side AC ($2x - y + 2 = 0$) and altitude CH ($2x - 3y + 4 = 0$). We have a system of linear equations:

1. $2x - y + 2 = 0 \quad \Rightarrow \quad y = 2x + 2$
2. $2x - 3y + 4 = 0$

Substitute $y = 2x+2$ from (1) into (2): $2x - 3(2x + 2) + 4 = 0$ $2x - 6x - 6 + 4 = 0$ $-4x - 2 = 0$ $-4x = 2 \quad \Rightarrow \quad x = -\frac{1}{2}$

Now, substitute $x = -\frac{1}{2}$ back into $y = 2x+2$: $y = 2\left(-\frac{1}{2}\right) + 2$ $y = -1 + 2 = 1$ So, the coordinates of vertex C are $\left(-\frac{1}{2}, 1\right)$.

• Common Mistake: Confusing orthocentre with other triangle centres (centroid, circumcentre, incenter). Remember the orthocentre is specifically about altitudes.

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4. Length of the Tangent from Point C to the Parabola

• Key Concept: The length of the tangent from an external point $(x_1, y_1)$ to the parabola $y^2 = 4ax$ is given by $\sqrt{S_1}$, where $S_1 = y_1^2 - 4ax_1$.
• Explanation: We have found the coordinates of point C. The parabola is given by $y^2 = 6x$. We need to evaluate the expression $y^2 - 6x$ at point C to find $S_1$, and then take its square root. The condition "in the first quadrant" means that the point of tangency (which is not C) lies in the first quadrant. This is usually to ensure a real tangent exists and to specify the branch of the parabola if it were symmetric (like $x^2=6y$). Here, $y^2=6x$ is symmetric about the x-axis, and the first quadrant part is for $y>0$.

The coordinates of point C are $\left(-\frac{1}{2}, 1\right)$. The equation of the parabola is $y^2 = 6x$, which can be written as $y^2 - 6x = 0$.

Let $S_1$ be the value of the expression $y^2 - 6x$ at point C$(x_C, y_C) = \left(-\frac{1}{2}, 1\right)$: $S_1 = y_C^2 - 6x_C$ $S_1 = (1)^2 - 6\left(-\frac{1}{2}\right)$ $S_1 = 1 - (-3)$ $S_1 = 1 + 3 = 4$

The length of the tangent, $L_T$, from C to the parabola is $\sqrt{S_1}$: $L_T = \sqrt{4}$ $L_T = 2$

• Tip: Always write the equation of the conic in the form $S=0$ (e.g., $y^2-6x=0$) before calculating $S_1$. If the equation was given as $y^2-6x+k=0$, then $S_1$ would be $y_1^2-6x_1+k$.
• Check: Since $S_1 = 4 > 0$, point C lies outside the parabola, and real tangents can be drawn. The parabola $y^2=6x$ opens to the right. $x_C = -1/2$ is negative, so C is indeed outside the parabola.

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Summary and Key Takeaway:

The length of the tangent from point C to the given parabola is 2.

This problem is an excellent example of how multiple concepts from coordinate geometry (lines, intersection of lines, orthocentre, slopes of perpendicular lines, and properties of parabolas) are integrated into a single question. The systematic approach of breaking down the problem into smaller, manageable steps is crucial for solving such complex problems accurately. Careful algebraic manipulation and understanding of geometric definitions are key to success.

The final answer is $\boxed{2}$.