Find the values of k for which the line (k–3) x – (4 – k²) y + k² –7k + 6 = 0 is

Passing through the origin.

The given equation of the

(k–3) x – (4 – k²) y + 6

If the given line is passing through the origin, then point (0,0) satisfies the given equation of line.

(k -3) (0) – (4-k²)(0) + k² -7k +6 = 0

⇒ k² -7k +6 = 0

⇒ k² -6k-k +6 = 0

⇒ (k-6)(k-1)=0

⇒ k = 1or 6

Therefore, if the given line is passing through the origin, then the value of k is either 1 or 6.