Let the direction cosines be \(l, m, n\). We are given that \(l = \sqrt{3}k\), \(m = \sqrt{3}k\), and \(n = \sqrt{3}k\).

We know that the sum of the squares of the direction cosines is equal to 1, i.e., \(l^2 + m^2 + n^2 = 1\).

Substituting the given values, we have:

\((\sqrt{3}k)^2 + (\sqrt{3}k)^2 + (\sqrt{3}k)^2 = 1\)

\(3k^2 + 3k^2 + 3k^2 = 1\)

\(9k^2 = 1\)

\(k^2 = \frac{1}{9}\)

Taking the square root of both sides, we get:

\(k = \pm\sqrt{\frac{1}{9}}\)

\(k = \pm\frac{1}{3}\)