The given pair of equations are kx + 3y − 1 = 0 and 12x + ky − 2 = 0
The pair of equations will have no solution if \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}\)
Where,
a1 = k, a2 = 12, b1 = 3, b2 = k, c1 = -1, c2 = -2
\(\therefore\) \(\frac k{12} = \frac 3k\)
⇒ \(k^2 = 36\)
⇒ \(k = \pm 6\)
\(\therefore\) \(\frac 3k\ne \frac{-1}{-2}\)
⇒ \(k\ne 6\)
\(\therefore\) \(k = -6\)
Hence, the value of k is −6.
