Answer: (i) \(a=2\).

(ii) The perpendicular distance from \(P\) to the plane is \(\sqrt{10}\).

(iii) The perpendicular distance from \(P\) to \(l_2\) is \(\sqrt{10}\).

A point on \(l_1\) is \((a+\lambda,9+2\lambda,13+3\lambda)\), and a point on \(l_2\) is \((-3-\mu,7+2\mu,-2-3\mu)\).

Since the lines intersect, compare coordinates. From the \(y\)-coordinates,

\(9+2\lambda=7+2\mu\), so \(\lambda=\mu-1\).

From the \(z\)-coordinates,

\(13+3\lambda=-2-3\mu\).

Substitute \(\lambda=\mu-1\):

\(13+3(\mu-1)=-2-3\mu\),

so \(10+3\mu=-2-3\mu\), hence \(\mu=-2\) and \(\lambda=-3\).

Using the \(x\)-coordinates,

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

so \(a-3=-1\), giving

\(a=2\).

The common point is obtained from \(l_2\) using \(\mu=-2\):

\(C=(-3,7,-2)+(-2)(-1,2,-3)=(-1,3,4).\)

The direction vectors of the two lines are

\(\mathbf d_1=(1,2,3),\qquad \mathbf d_2=(-1,2,-3).\)

A normal to the plane containing both lines is

\(\mathbf n=\mathbf d_1\times\mathbf d_2=(-12,0,4),\)

so we may use \(\mathbf n=(-3,0,1)\). The plane through \(C=(-1,3,4)\) is

\(-3(x+1)+(z-4)=0,\)

or

\(-3x+z-7=0.\)

For \(P=(3,1,6)\), the perpendicular distance to this plane is

\(\dfrac{| -3(3)+6-7 |}{\sqrt{(-3)^2+1^2}}=\dfrac{10}{\sqrt{10}}=\sqrt{10}.\)

For the distance from \(P\) to \(l_2\), take \(B=(-3,7,-2)\) on \(l_2\). Then

\(\overrightarrow{BP}=P-B=(6,-6,8).\)

The distance from a point to a line is

\(\dfrac{|\overrightarrow{BP}\times\mathbf d_2|}{|\mathbf d_2|}.\)

Now

\(\overrightarrow{BP}\times\mathbf d_2=(6,-6,8)\times(-1,2,-3)=(2,10,6),\)

so

\(|\overrightarrow{BP}\times\mathbf d_2|=\sqrt{2^2+10^2+6^2}=2\sqrt{35},\qquad |\mathbf d_2|=\sqrt{14}.\)

Therefore the distance is

\(\dfrac{2\sqrt{35}}{\sqrt{14}}=\sqrt{10}.\)