1. Interpret the three lines correctly

The first part of the statement is clearly the pair of lines $4x+3y=69, \qquad 4y-3x=17, \qquad x+7y=61.$ So the triangle is formed by these three lines.

Let us denote them as:

• $L_1: 4x+3y=69$
• $L_2: -3x+4y=17$
• $L_3: x+7y=61$

We first find the three vertices of the triangle.

2. Find vertex $A=L_1\cap L_2$

Solve $4x+3y=69 \quad ...(1)$ $-3x+4y=17 \quad ...(2)$

Multiply (1) by $4$ and (2) by $3$: $16x+12y=276$ $-9x+12y=51$ Subtract: $25x=225 \Rightarrow x=9$ Then from (1): $4(9)+3y=69$ $36+3y=69$ $3y=33 \Rightarrow y=11$ So, $A=(9,11).$

3. Find vertex $B=L_2\cap L_3$

Solve $-3x+4y=17 \quad ...(2)$ $x+7y=61 \quad ...(3)$

From (3): $x=61-7y$ Substitute into (2): $-3(61-7y)+4y=17$ $-183+21y+4y=17$ $25y=200 \Rightarrow y=8$ Then $x=61-56=5$ So, $B=(5,8).$

4. Find vertex $C=L_1\cap L_3$

Solve $4x+3y=69 \quad ...(1)$ $x+7y=61 \quad ...(3)$

From (3): $x=61-7y$ Substitute into (1): $4(61-7y)+3y=69$ $244-28y+3y=69$ $244-25y=69$ $25y=175 \Rightarrow y=7$ Then $x=61-49=12$ So, $C=(12,7).$

Thus the triangle has vertices $A=(9,11),\quad B=(5,8),\quad C=(12,7).$

5. Check whether the triangle is right-angled

Compute squared side lengths:

$AB^2=(9-5)^2+(11-8)^2=4^2+3^2=25$ $AC^2=(12-9)^2+(7-11)^2=3^2+(-4)^2=25$ $BC^2=(12-5)^2+(7-8)^2=7^2+(-1)^2=50$

Since $AB^2+AC^2=25+25=50=BC^2,$ triangle $ABC$ is right-angled at $A$.

For a right triangle, the circumcenter is the midpoint of the hypotenuse $BC$.

So circumcenter $(\alpha,\beta)$ is midpoint of $(5,8)$ and $(12,7)$: $\alpha=\frac{5+12}{2}=\frac{17}{2}, \qquad \beta=\frac{8+7}{2}=\frac{15}{2}.$

6. Compute the required value

We need $(\alpha-\beta)^2+\alpha+\beta.$

Now, $\alpha-\beta=\frac{17}{2}-\frac{15}{2}=1$ so $(\alpha-\beta)^2=1.$ Also, $\alpha+\beta=\frac{17}{2}+\frac{15}{2}=16.$ Therefore, $(\alpha-\beta)^2+\alpha+\beta=1+16=17.$

7. Option check

The value is $\boxed{17}$ which corresponds to Option B.