1. Find the center of the circle

Since the given lines are diameters, both pass through the center of the circle. So the center is the intersection of $2x-3y=5$ and $3x-4y=7.$

Solve:

From $2x-3y=5 \,\quad ...(1)$ $3x-4y=7 \,\quad ...(2)$

Multiply (1) by $3$ and (2) by $2$: $6x-9y=15$ $6x-8y=14$

Subtract: $-y=1 \Rightarrow y=-1.$

Substitute into (1): $2x-3(-1)=5$ $2x+3=5$ $2x=2$ $x=1.$

Hence, center is $C(1,-1).$

2. Find the equation of the line joining the given points

The points are $A\left(-\frac{22}{7},-4\right), \quad B\left(-\frac{1}{7},3\right).$

Slope: $m=\frac{3-(-4)}{-\frac17-\left(-\frac{22}{7}\right)}=\frac{7}{\frac{21}{7}}=\frac{7}{3}.$

So equation through $B\left(-\frac17,3\right)$ is $y-3=\frac{7}{3}\left(x+\frac17\right).$

Simplify: $3y-9=7x+1$ $7x-3y+10=0.$

3. Why does this line intersect the circle at only one point?

A line meeting a circle at only one point is a tangent. So the given line is tangent to the circle at point $P$.

The radius to the point of contact is perpendicular to the tangent. Thus, $P$ is the foot of the perpendicular from the center $C(1,-1)$ to the line $7x-3y+10=0.$

4. Find the foot of perpendicular from $(1,-1)$ to $7x-3y+10=0$

For line $ax+by+c=0,$ foot of perpendicular from $(x_1,y_1)$ is $\left(x_1-\frac{a(ax_1+by_1+c)}{a^2+b^2},\; y_1-\frac{b(ax_1+by_1+c)}{a^2+b^2}\right).$

Here, $a=7,\; b=-3,\; c=10,\; (x_1,y_1)=(1,-1).$

First compute: $ax_1+by_1+c=7(1)+(-3)(-1)+10=7+3+10=20.$

Also, $a^2+b^2=49+9=58.$

So $\alpha=1-\frac{7\cdot 20}{58}=1-\frac{140}{58}=1-\frac{70}{29}=\frac{29-70}{29}=-\frac{41}{29},$ $\beta=-1-\frac{-3\cdot 20}{58}=-1+\frac{60}{58}=-1+\frac{30}{29}=\frac{-29+30}{29}=\frac{1}{29}.$

Thus, $P\left(-\frac{41}{29},\frac{1}{29}\right).$

5. Compute $17\beta-\alpha$