$$ \text{Let the lines } 3x-4y-\alpha=0,\; 8x-11y-33=0,\; \text{ and } 2x-3y+\lambda=0$$ $$\text{ be concurrent. If the image of the point } (1,2) \text{ in the line } 2x-3y+\lambda=0 $$ $$\text{ is } \left(\frac{57}{13},-\frac{40}{13}\right), \text{ then } |\alpha\lambda| \text{ is equal to:} $$

Lines $$3x-4y-\alpha=0$$, $$8x-11y-33=0$$, and $$2x-3y+\lambda=0$$ are concurrent, and the image of (1,2) in the line $$2x-3y+\lambda=0$$ is $$(\frac{57}{13}, -\frac{40}{13}).$$

The midpoint of (1,2) and $$(\frac{57}{13}, -\frac{40}{13})$$ lies on $$2x-3y+\lambda=0$$, and is calculated as $$\left(\frac{1+\frac{57}{13}}{2}, \frac{2-\frac{40}{13}}{2}\right)=\left(\frac{\frac{70}{13}}{2}, \frac{-\frac{14}{13}}{2}\right)=\left(\frac{35}{13}, -\frac{7}{13}\right).$$ Substituting this into the line gives
$$2\left(\frac{35}{13}\right)-3\left(-\frac{7}{13}\right)+\lambda=0\quad\Longrightarrow\quad\frac{70}{13}+\frac{21}{13}+\lambda=0\quad\Longrightarrow\quad7+\lambda=0\quad\Longrightarrow\quad\lambda=-7.$$

The condition for concurrency of the three lines is
$$\begin{vmatrix}3 & -4 & -\alpha\\8 & -11 & -33\\2 & -3 & -7\end{vmatrix}=0.$$ Expanding, we obtain
$$3\bigl((-11)(-7)-(-33)(-3)\bigr)-(-4)\bigl((8)(-7)-(-33)(2)\bigr)+(-\alpha)\bigl((8)(-3)-(-11)(2)\bigr)$$
$$=3(77-99)+4(-56+66)+(-\alpha)(-24+22)$$
$$=3(-22)+4(10)+2\alpha$$
$$=-66+40+2\alpha=-26+2\alpha=0\quad\Longrightarrow\quad\alpha=13.$$

Hence
$$|\alpha\lambda|=|13\times(-7)|=91.$$ The correct answer is Option 3: 91.