Let the equations of two sides of a triangle be $$3x - 2y + 6 = 0$$ and $$4x + 5y - 20 = 0$$. If the orthocenter of this triangle is at $$(1, 1)$$ then the equation of its third side is:

We have two sides of the triangle given by the straight-line equations $$3x-2y+6=0 \quad\text{and}\quad 4x+5y-20=0.$$ Their point of intersection will be one vertex of the triangle. Let us find that vertex first.

Solving the simultaneous equations $$\begin{aligned} 3x-2y+6&=0,\\ 4x+5y-20&=0, \end{aligned}$$ we get $$\begin{aligned} 3x-2y&=-6,\\ 4x+5y&=20. \end{aligned}$$ Multiply the first by $$5$$ and the second by $$2$$: $$\begin{aligned} 15x-10y&=-30,\\ 8x+10y&=40. \end{aligned}$$ Adding, $$23x=10 \;\Longrightarrow\; x=\dfrac{10}{23}.$$ Putting this in $$3x-2y=-6$$, $$3\!\left(\dfrac{10}{23}\right)-2y=-6 \;\Longrightarrow\; \dfrac{30}{23}-2y=-6 \;\Longrightarrow\; -2y=-6-\dfrac{30}{23} =-\dfrac{168}{23} \;\Longrightarrow\; y=\dfrac{84}{23}.$$ Hence the vertex at the intersection of the two given sides is $$C\!\left(\dfrac{10}{23},\,\dfrac{84}{23}\right).$$

The orthocenter is given as $$H(1,1).$$ Recall that “the orthocenter is the common point of the three altitudes of a triangle.” Therefore, the altitude from the unknown vertex $$B$$ (lying on the first side) is perpendicular to the second side, and the altitude from the unknown vertex $$A$$ (lying on the second side) is perpendicular to the first side.

Altitude from $$B$$ to the side $$4x+5y-20=0$$ For the line $$4x+5y-20=0$$ the normal (perpendicular) vector is $$(4,5)$$. Hence a line that is perpendicular to this side must have the same direction vector $$(4,5)$$ and therefore slope $$m=\dfrac{5}{4}.$$ Because the altitude passes through the orthocenter $$H(1,1)$$ and the vertex $$B(x_b,y_b)$$, the slope condition gives $$\dfrac{y_b-1}{x_b-1}=\dfrac{5}{4}\;\Longrightarrow\;4(y_b-1)=5(x_b-1) \;\Longrightarrow\;5x_b-4y_b=1. \quad -(1)$$ At the same time, $$B$$ lies on the first side $$3x-2y+6=0$$: $$3x_b-2y_b+6=0 \;\Longrightarrow\; 3x_b-2y_b=-6. \quad -(2)$$ Solving (1) and (2): from (2) $$x_b=\dfrac{-6+2y_b}{3},$$ substitute into (1): $$5\!\left(\dfrac{-6+2y_b}{3}\right)-4y_b=1 \;\Longrightarrow\; \dfrac{-30+10y_b}{3}-4y_b=1 \;\Longrightarrow\; -30+10y_b-12y_b=3 \;\Longrightarrow\; -2y_b=33 \;\Longrightarrow\; y_b=-\dfrac{33}{2},$$ and hence $$x_b=\dfrac{-6+2\!\left(-\dfrac{33}{2}\right)}{3}=\dfrac{-6-33}{3}=-13.$$ Thus $$B(-13,-33/2).$$

Altitude from $$A$$ to the side $$3x-2y+6=0$$ For the line $$3x-2y+6=0$$ the normal vector is $$(3,-2)$$, so a perpendicular line has that direction and hence slope $$m=\dfrac{-2}{3}.$$ Because the altitude passes through $$H(1,1)$$ and $$A(x_a,y_a)$$, $$\dfrac{y_a-1}{x_a-1}=-\dfrac{2}{3} \;\Longrightarrow\;3(y_a-1)=-2(x_a-1) \;\Longrightarrow\;2x_a+3y_a=5. \quad -(3)$$ Also $$A$$ lies on the second side $$4x+5y-20=0$$: $$4x_a+5y_a-20=0 \;\Longrightarrow\;4x_a+5y_a=20. \quad -(4)$$ Solving (3) and (4): multiply (3) by $$2$$: $$4x_a+6y_a=10$$. Subtract this from (4): $$(4x_a+5y_a)-(4x_a+6y_a)=20-10 \;\Longrightarrow\;-y_a=10 \;\Longrightarrow\; y_a=-10,$$ then from (3) $$2x_a+3(-10)=5 \;\Longrightarrow\; 2x_a=35 \;\Longrightarrow\; x_a=\dfrac{35}{2}.$$ So $$A\!\left(\dfrac{35}{2},-10\right).$$

The third side is the straight line through $$A$$ and $$B$$. Using the two-point form, for points $$\bigl(x_1,y_1\bigr)=A$$ and $$\bigl(x_2,y_2\bigr)=B$$, $$(y-y_1)(x_2-x_1)=(y_2-y_1)(x-x_1).$$ Here $$x_2-x_1=-13-\dfrac{35}{2}=-\dfrac{61}{2},\qquad y_2-y_1=-\dfrac{33}{2}-(-10)=-\dfrac{13}{2}.$$ Substituting, $$(y+10)\!\left(-\dfrac{61}{2}\right)=\!\left(-\dfrac{13}{2}\right)\!\left(x-\dfrac{35}{2}\right).$$ Multiply by $$2$$ to clear the denominators: $$-61(y+10)=-13\!\left(x-\dfrac{35}{2}\right).$$ Expand both sides: $$-61y-610=-13x+\dfrac{455}{2}.$$ Multiply by $$2$$ once more to eliminate the remaining fraction: $$-122y-1220=-26x+455.$$ Bring every term to the left: $$26x-122y-1675=0.$$

This matches option B in the list. Hence, the correct answer is Option 2.