Answer:

(i) For the two lines to intersect, there must be values of \(\lambda\) and \(\mu\) giving the same point.

So equate components:

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

From the first equation, \(\mu=2\lambda\). Substituting into the second gives \(-2+\lambda=-5-2\lambda\), so \(3\lambda=-3\) and \(\lambda=-1\). Hence \(\mu=-2\).

Check in the third equation: \(-4(-1)=4\) and \(2-(-2)=4\), so it is satisfied. Therefore the lines intersect.

(ii) The plane containing the two intersecting lines has a normal vector given by the cross product of their direction vectors.

The direction vectors are \(\mathbf{a}=(2,1,-4)\) and \(\mathbf{b}=(1,-1,-1)\).

So

\(\mathbf{n}=\mathbf{a}\times\mathbf{b}=(2,1,-4)\times(1,-1,-1)=(-5,-2,-3)\).

Using the point on \(l_1\) with \(\lambda=0\), namely \((4,-2,0)\), the plane equation is

\(-5(x-4)-2(y+2)-3z=0\),

which simplifies to \(5x+2y+3z=16\).

For \(P=(3,-5,6)\), the perpendicular distance to this plane is

\(\frac{|5(3)+2(-5)+3(6)-16|}{\sqrt{5^2+2^2+3^2}}=\frac{|15-10+18-16|}{\sqrt{38}}=\frac{7}{\sqrt{38}}.\)

(iii) A point on \(l_1\) is \(A=(4,-2,0)\), and the direction vector of \(l_1\) is \((2,1,-4)\).

First find \(\overrightarrow{AP} = (3,-5,6)-(4,-2,0)=(-1,-3,6)\).

The perpendicular distance from \(P\) to \(l_1\) is

\(\frac{|\overrightarrow{AP}\times(2,1,-4)|}{|(2,1,-4)|}.\)

Now

\(\overrightarrow{AP}\times(2,1,-4)=(-1,-3,6)\times(2,1,-4)=(6,8,5).\)

Hence

\(|\overrightarrow{AP}\times(2,1,-4)|=\sqrt{6^2+8^2+5^2}=\sqrt{125},\qquad |(2,1,-4)|=\sqrt{21}.\)

So the distance is

\(\frac{\sqrt{125}}{\sqrt{21}}=\sqrt{\frac{125}{21}}=\frac{5\sqrt{105}}{21}.\)

(i) Write the two lines in component form:

\(l_1: (x,y,z)=(4+2\lambda,-2+\lambda,-4\lambda)\)

\(l_2: (x,y,z)=(4+\mu,-5-\mu,2-\mu)\)

For an intersection we need the same point, so

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

From the first equation, \(\mu=2\lambda\). Substituting into the second gives

\(-2+\lambda=-5-2\lambda\).

So \(3\lambda=-3\), hence \(\lambda=-1\), and therefore \(\mu=-2\).

Check the third equation:

\(-4(-1)=4\) and \(2-(-2)=4\).

This is consistent, so the lines intersect.

(ii) The direction vectors of the lines are

\(\mathbf{a}=(2,1,-4)\), \(\mathbf{b}=(1,-1,-1)\).

A normal to the plane containing both lines is

\(\mathbf{n}=\mathbf{a}\times\mathbf{b}=(2,1,-4)\times(1,-1,-1)=(-5,-2,-3).\)

Using the point \((4,-2,0)\) on \(l_1\), the plane equation is

\(-5(x-4)-2(y+2)-3z=0.\)

Expanding gives

\(-5x+20-2y-4-3z=0,\)

so

\(5x+2y+3z=16.\)

The perpendicular distance from \(P=(3,-5,6)\) to this plane is

\(\frac{|5(3)+2(-5)+3(6)-16|}{\sqrt{5^2+2^2+3^2}}=\frac{|15-10+18-16|}{\sqrt{38}}=\frac{7}{\sqrt{38}}.\)

(iii) Let \(A=(4,-2,0)\) be a point on \(l_1\). Then

\(\overrightarrow{AP}=(3,-5,6)-(4,-2,0)=(-1,-3,6).\)

The shortest distance from \(P\) to the line \(l_1\) is

\(\frac{|\overrightarrow{AP}\times(2,1,-4)|}{|(2,1,-4)|}.\)

Now

\(\overrightarrow{AP}\times(2,1,-4)=(-1,-3,6)\times(2,1,-4)=(6,8,5).\)

So

\(|\overrightarrow{AP}\times(2,1,-4)|=\sqrt{6^2+8^2+5^2}=\sqrt{125},\qquad |(2,1,-4)|=\sqrt{21}.\)

Therefore

\(d=\frac{\sqrt{125}}{\sqrt{21}}=\sqrt{\frac{125}{21}}=\frac{5\sqrt{105}}{21}.\)