Using distance formula, $PQ = \sqrt{(q-p)^2 + 0^2} = q-p$ and $RS = \sqrt{(0-0)^2 + (q-p)^2} = q-p$. The sides are equal and diagonals are not equal, thus it is a parallelogram.

If the origin is shifted to $(h, k)$, the new coordinates are $(X, Y) = (x-h, y-k)$. Here $(h, k) = (2, 2)$ and $(x, y) = (4, -2)$. Therefore, $X = 4 - 2 = 2$ and $Y = -2 - 2 = -4$. The new coordinates are $(2, -4)$.

Slope ($m$) = $\frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - 7}{-2 - 3} = \frac{-6}{-5} = \frac{6}{5}$.

In the third quadrant, both the abscissa ($x$-coordinate) and the ordinate ($y$-coordinate) are negative.

Since the points $(a, 0)$, $(0, b)$ and $(1, 1)$ are collinear, the area of the triangle formed by them will be zero: $\frac{1}{2} |a(b-1) + 0(1-0) + 1(0-b)| = 0 \implies ab - a - b = 0 \implies ab = a + b$. Dividing both sides by $ab$, $1 = \frac{1}{b} + \frac{1}{a}$.

For the pair of lines $x^2 - 4xy + y^2 = 0$, using the angle formula $\tan \theta = \frac{2\sqrt{h^2 - ab}}{a + b}$, the angle between lines is $60^{\circ}$. The angles with the third line are also $60^{\circ}$, thus it is an equilateral triangle.

Let the ratio be $k:1$. Formula: $x = \frac{m x_2 + n x_1}{m + n}$. $-3 = \frac{k(-2) + 1(-5)}{k+1} \implies -3k - 3 = -2k - 5 \implies -3+5 = 3k-2k \implies 2 = k$. Thus, the ratio is $2:1$ (Note: There is a discrepancy in the provided options; calculation yields $2:1$).

Point P divides the line segment in the ratio 1:2. Coordinates of P: $P = (\frac{1\cdot5 + 2\cdot2}{1+2}, \frac{1(-8) + 2(1)}{1+2}) = (\frac{9}{3}, \frac{-6}{3}) = (3, -2)$. Point P lies on $2x - y + k = 0$, so $2(3) - (-2) + k = 0 \implies 6 + 2 + k = 0 \implies 8 + k = 0 \implies k = -8$.

The equation of the circle with endpoints of the diameter as $(x_1, y_1)$ and $(x_2, y_2)$ is given by $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$. Here $(1, 2)$ and $(3, 4)$ are given. Therefore, $(x - 1)(x - 3) + (y - 2)(y - 4) = 0 \implies (x^2 - 4x + 3) + (y^2 - 6y + 8) = 0 \implies x^2 + y^2 - 4x - 6y + 11 = 0$.

The centroid is at the origin $(0, 0)$, so $\frac{a+b+c}{3} = 0 \implies a+b+c = 0$. We know that if $a+b+c = 0$, then $a^3 + b^3 + c^3 = 3abc$. The mean of $a^3, b^3, c^3$ is $\frac{a^3 + b^3 + c^3}{3} = \frac{3abc}{3} = abc$.

Let circumcenter be $O(x, y)$. $OA^2 = OB^2 = OC^2$. Equating distances, $(x+2)^2 + (y+3)^2 = (x+1)^2 + y^2 \implies x^2+4x+4+y^2+6y+9 = x^2+2x+1+y^2 \implies 2x + 6y = -12 \implies x + 3y = -6$. And $(x+1)^2 + y^2 = (x-7)^2 + (y+6)^2 \implies x^2+2x+1+y^2 = x^2-14x+49+y^2+12y+36 \implies 16x - 12y = 84 \implies 4x - 3y = 21$. Solving both equations gives $5x = 15 \implies x = 3$. Then $3 + 3y = -6 \implies 3y = -9 \implies y = -3$. Thus, the circumcenter is $(3, -3)$.

The abscissa ($x$-coordinate) is negative in the second and third quadrants. (Note: The standard answer is II and III).

Let the ratio be $k:1$. Using the section formula: $x = \frac{m x_2 + n x_1}{m + n}$. Here, $-4 = \frac{k(3) + 1(-6)}{k+1}$. Solving this: $-4k - 4 = 3k - 6 \implies 7k = 2 \implies k = \frac{2}{7}$. Thus, the ratio is $2:7$.

Circle 1 has center $(1, 3)$ and radius $r$. Circle 2 is $x^2 + y^2 - 8x + 2y + 8 = 0$, with center $(4, -1)$ and radius $R = \sqrt{4^2 + (-1)^2 - 8} = \sqrt{16+1-8} = 3$. Distance between centers $d = \sqrt{(4-1)^2 + (-1-3)^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9+16} = 5$. Two circles intersect if $|R - r| < d < R + r$. Therefore, $|3 - r| < 5 < 3 + r$. This yields $r > 2$ and $r < 8$, i.e., $2 < r < 8$.

In the second quadrant, the value of $x$ is negative and the value of $y$ is positive. Therefore, $(-2, 3)$ lies in the second quadrant.

Any point on the X-axis has a y-coordinate of zero. Therefore, the point will be of the form $(a, 0)$. Among the options, $(2, 0)$ is correct.

In the Cartesian plane, the coordinates of the origin are always $(0, 0)$.

The distance of point $(x, y)$ from the y-axis is $|x|$ and from the x-axis is $|y|$. Therefore, the sum of both distances is $|x| + |y|$.

The coordinates of a point on the y-axis are $(0, a)$. The distance is equal: $\sqrt{(7-0)^2 + (6-a)^2} = \sqrt{(-3-0)^2 + (4-a)^2}$. Solving: $49 + 36 + a^2 - 12a = 9 + 16 + a^2 - 8a \implies 85 - 12a = 25 - 8a \implies 60 = 4a \implies a = 15$.

The midpoint is $(x, y) = (\frac{3+P}{2}, \frac{4+7}{2}) = (\frac{3+P}{2}, \frac{11}{2})$. It satisfies the equation $2x + 2y + 1 = 0$. Substituting the values: $2(\frac{3+P}{2}) + 2(\frac{11}{2}) + 1 = 0 \implies 3+P + 11 + 1 = 0 \implies P + 15 = 0 \implies P = -15$.