Given equation of a line is
(k – 3)x – (4 – k2) y + k2 – 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 – k2) = 0
(Since Any line parallel to y-axis is of the form x = a)
⇒ k2 = 4, i.e. k = ± 2.
(c) Since the line (i) passes through the origin if (0, 0) lies on (i)
⇒ 0 – 0 + k2 – 7k + 6 = 0
⇒ (k – 6) (k – 1) = 0, i.e. k = 6, 1.