Express the general point on the line l in component form:

\((1 + 2\lambda, 2 - \lambda, 1 + \lambda)\).

The distance from the origin is given by:

\(\sqrt{(1 + 2\lambda)^2 + (2 - \lambda)^2 + (1 + \lambda)^2} = \sqrt{10}\).

Squaring both sides, we have:

\((1 + 2\lambda)^2 + (2 - \lambda)^2 + (1 + \lambda)^2 = 10\).

Expanding and simplifying:

\(1 + 4\lambda + 4\lambda^2 + 4 - 4\lambda + \lambda^2 + 1 + 2\lambda + \lambda^2 = 10\).

Combine like terms:

\(6\lambda^2 + 2\lambda + 6 = 10\).

Rearrange to form a quadratic equation:

\(6\lambda^2 + 2\lambda - 4 = 0\).

Divide the entire equation by 2:

\(3\lambda^2 + \lambda - 2 = 0\).

Using the quadratic formula \(\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3\), \(b = 1\), \(c = -2\):

\(\lambda = \frac{-1 \pm \sqrt{1 + 24}}{6} = \frac{-1 \pm 5}{6}\).

This gives \(\lambda = \frac{2}{3}\) or \(\lambda = -1\).

Substitute back to find the position vectors:

For \(\lambda = -1\):

\((1 + 2(-1), 2 - (-1), 1 + (-1)) = (-1, 3, 0)\).

For \(\lambda = \frac{2}{3}\):

\((1 + 2\left(\frac{2}{3}\right), 2 - \left(\frac{2}{3}\right), 1 + \left(\frac{2}{3}\right)) = \left(\frac{7}{3}, \frac{4}{3}, \frac{5}{3}\right)\).