Answer: The plane is \(5x-3y-2z=7\).

The foot of the perpendicular from \(D(1,10,3)\) to the plane is \((6,7,1)\).

The shortest distance between \(l_1\) and \(l_3\) is

\(\frac{10}{\sqrt{42}}=\frac{5\sqrt{42}}{21}.\)

The two lines \(l_1\) and \(l_2\) pass through \(A=(6,5,4)\) and have direction vectors

\(\mathbf d_1=(1,1,1),\qquad \mathbf d_2=(4,6,1).\)

A normal vector to the plane is

\(\mathbf d_1\times\mathbf d_2=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\1&1&1\\4&6&1\end{vmatrix}=(-5,3,2).\)

Using the opposite normal \((5,-3,-2)\), the plane has equation

\(5x-3y-2z=c.\)

Substitute \(A=(6,5,4)\):

\(c=5(6)-3(5)-2(4)=30-15-8=7.\)

So

\(5x-3y-2z=7.\)

The line through \(D=(1,10,3)\) perpendicular to the plane has direction equal to the normal vector:

\((x,y,z)=(1,10,3)+t(5,-3,-2).\)

Substitute in the plane:

\(5(1+5t)-3(10-3t)-2(3-2t)=7.\)

This gives

\(5+25t-30+9t-6+4t=7,\)

\(38t-31=7,\qquad t=1.\)

Therefore the foot of the perpendicular is

\((1,10,3)+(5,-3,-2)=(6,7,1).\)

For the shortest distance between \(l_1\) and \(l_3\), use

\(d=\frac{|(\mathbf b-\mathbf a)\cdot(\mathbf u\times\mathbf v)|}{|\mathbf u\times\mathbf v|}.\)

Take \(\mathbf a=(6,5,4)\), \(\mathbf u=(1,1,1)\) for \(l_1\), and \(\mathbf b=(1,10,3)\), \(\mathbf v=(2,-3,1)\) for \(l_3\). Then

\(\mathbf u\times\mathbf v=(1,1,1)\times(2,-3,1)=(4,1,-5).\)

Also

\(\mathbf b-\mathbf a=(-5,5,-1).\)

So

\((\mathbf b-\mathbf a)\cdot(\mathbf u\times\mathbf v)=-20+5+5=-10.\)

Thus

\(d=\frac{10}{\sqrt{4^2+1^2+(-5)^2}}=\frac{10}{\sqrt{42}}=\frac{5\sqrt{42}}{21}.\)