Given equation of a line is (k – 3)x – (4 – $k^2$) y + $k^2$ – 7k + 6 = 0 ... (i) (a) The line (i) is parallel to x-axis if coefficient of x = 0 ⇒ k – 3 = 0 ⇒ k = 3 (Since Any line parallel to x-axis is of the form y = b) (b) Clearly the line (i) is parallel to y-axis if coefficient of y = 0 ⇒ (4 – $k^2$) = 0 (Since Any line parallel to y-axis is of the form x = a) ⇒ $k^2$ = 4, i.e. k = ± 2. (c) Since the line (i) passes through the origin if (0, 0) lies on (i) ⇒ 0 – 0 + $k^2$ – 7k + 6 = 0 ⇒ (k – 6) (k – 1) = 0, i.e. k = 6, 1.