Answer: (i)(a) \(a=-\frac{6}{13}\).

(i)(b) The plane is \(-13x+5y-z=43\).

(ii) \(a=3\).

(i)(a) A point on \(l_1\) is

\((\lambda a,\,9-\lambda,\,2+\lambda)\),

and a point on \(l_2\) is

\((-6-\mu a,\,-5+2\mu,\,10+4\mu)\).

If the lines intersect, then

\((\lambda a,9-\lambda,2+\lambda)=(-6-\mu a,-5+2\mu,10+4\mu).\)

From the second and third components,

\(9-\lambda=-5+2\mu,\qquad 2+\lambda=10+4\mu.\)

These give \(\lambda=12\) and \(\mu=1\). Substituting in the first component gives

\(12a=-6-a\),

so

\(\boxed{a=-\frac{6}{13}}.\)

(i)(b) With \(a=-\frac{6}{13}\), the direction vectors are

\(\mathbf{d}_1=\left(-\frac{6}{13},-1,1\right),\qquad \mathbf{d}_2=\left(\frac{6}{13},2,4\right).\)

A normal vector to the plane is

\(\mathbf{d}_1\times\mathbf{d}_2=\left(-6,\frac{30}{13},-\frac{6}{13}\right),\)

which is proportional to \((-13,5,-1)\). Using the point \((0,9,2)\),

\(-13(x-0)+5(y-9)-(z-2)=0.\)

Therefore the plane is

\(\boxed{-13x+5y-z=43}.\)

(ii) For the general value of \(a\),

\(\mathbf{d}_1=(a,-1,1),\qquad \mathbf{d}_2=(-a,2,4),\)

so

\(\mathbf{d}_1\times\mathbf{d}_2=(-6,-5a,a).\)

Hence

\(|\mathbf{d}_1\times\mathbf{d}_2|=\sqrt{36+26a^2}.\)

The vector between the given points on the two lines is

\((-6,-5,10)-(0,9,2)=(-6,-14,8).\)

Therefore

\((-6,-14,8)\cdot(-6,-5a,a)=36+78a.\)

The distance between the skew lines is

\(\frac{|36+78a|}{\sqrt{36+26a^2}}.\)

Since the distance is \(3\sqrt{30}\),

\(3\sqrt{30}\sqrt{36+26a^2}=|36+78a|.\)

Squaring and simplifying gives

\(15(18+13a^2)=(6+13a)^2.\)

Thus

\(26(a-3)^2=0\),

so

\(\boxed{a=3}.\)