(i) To show that the lines intersect, equate the vector equations:

\(1 + \lambda = a + \mu\)

\(4 = 2 + 2\mu\)

\(-2 + 3\lambda = -2 + 3a\mu\)

From the second equation, solve for \(\mu\):

\(2 = 2\mu \Rightarrow \mu = 1\)

Substitute \(\mu = 1\) into the first equation:

\(1 + \lambda = a + 1 \Rightarrow \lambda = a\)

Check the third equation:

\(-2 + 3a = -2 + 3a\)

All equations are satisfied, so the lines intersect for all \(a\).

(ii) The point of intersection is \((a + 1, 4, 3a - 2)\).

The distance from the origin is given by:

\(\sqrt{(a+1)^2 + 4^2 + (3a-2)^2} = 9\)

\((a+1)^2 + 16 + (3a-2)^2 = 81\)

\(a^2 + 2a + 1 + 16 + 9a^2 - 12a + 4 = 81\)

\(10a^2 - 10a + 21 = 81\)

\(10a^2 - 10a - 60 = 0\)

\(a^2 - a - 6 = 0\)

Factor the quadratic:

\((a - 3)(a + 2) = 0\)

Thus, \(a = 3\) or \(a = -2\).