Concept: The equation of tangent at any point (h, k) on the parabola y^2 = 8x is given by y = mx - 2am - k, where m is the slope of the tangent and a is the distance of the point (h, k) from the focus (4, 0).

Given: The equation of the line is x - y + 3 = 0.

Approach:
1. Find the points of intersection of the line and the parabola.
2. Find the equation of tangent at each point of intersection.
3. Find the coordinates of the point of intersection of the variable chords of contact.
4. Show that the point obtained in step 3 is fixed.

Solution:
1. The points of intersection of the line and the parabola are obtained by solving the simultaneous equations x - y + 3 = 0 and y^2 = 8x. Substituting x = y - 3 in the second equation, we get (y - 2)^2 = 4. Hence, y = 2 ± 2, i.e. y = 4 or y = 0. Therefore, the points of intersection are (1, 4) and (9, 0).

2. The slope of tangent at any point (h, k) on the parabola y^2 = 8x is given by m = 2a/k, where a = 2 and k = (h^2)/8. Therefore, the equation of tangent at (h, k) is y = mx - 2am - k = (2a/h)x - h/2.

At (1, 4), the equation of tangent is y = (4/3)x - 5/2.
At (9, 0), the equation of tangent is y = (4/27)x - 27/2.

3. Let (x1, y1) and (x2, y2) be the points of intersection of the tangents at (1, 4) and (9, 0), respectively. Then, x1 - y1 + 3 = 0 and y1^2 = 8x1, and x2 - y2 + 3 = 0 and y2^2 = 8x2. Eliminating x1 and x2, we get y1^2y2^2 = 64(y1 + 3)(y2 + 3). Therefore, the coordinates of the point of intersection are (3, 4).

4. To show that the point obtained in step 3 is fixed, we need to show that it is independent of the choice of points on the line x - y + 3 = 0. Let (a, a + 3) be any point on the line. Then, the equation of tangent at (a/2, a^2/8 + 3/2) is y = (a/4)x - (a^2/4) - 3/2. The point of intersection of the tangents at (1, 4) and (a/2, a^2/8 + 3/2) is given by (2a/3, 2a^2/3