Let the common ratio be \(r\).

\[ x_1 = a, \; x_2 = ar, \; x_3 = ar^2 \] \[ y_1 = b, \; y_2 = br, \; y_3 = br^2 \]

So the points are \((a,b), \; (ar,br), \; (ar^2,br^2)\).

Slopes:

Between first two points: \[ m_{12} = \frac{br - b}{ar - a} = \frac{b(r-1)}{a(r-1)} = \frac{b}{a} \] Between second and third points: \[ m_{23} = \frac{br^2 - br}{ar^2 - ar} = \frac{br(r-1)}{ar(r-1)} = \frac{b}{a} \]

Since \(m_{12} = m_{23}\), the points are collinear.

Final Answer: The points \((x_1,y_1), (x_2,y_2), (x_3,y_3)\) are collinear.

Step 1: Slope of line through points (-1, 4) and (0, 6).
$$m = \frac{6 - 4}{0 - (-1)} = \frac{2}{1} = 2$$

Step 2: For parallelism, slope of line through (3, y) and (2, 7) must also be 2.
$$\frac{y - 7}{3 - 2} = 2$$

Step 3: Solve for y.
$$y - 7 = 2(1)$$ $$y - 7 = 2$$ $$y = 9$$

Correct Value of y: 9

Answer: Option 2