Answer: (i) \(k=11\) or \(k=3\). (ii) For \(k=11\): tangent \(y=-5x+16\), curve \(y=7-11x-x^2\), point \((-3,31)\). For \(k=3\): tangent \(y=-5x+8\), curve \(y=7-3x-x^2\), point \((1,3)\). (iii) Distance \(=\sqrt{800}=20\sqrt2\).

(i) At a point of contact, the line and curve meet in one repeated root. Equate the two expressions for \(y\):

\(-5x+k+5=7-kx-x^2.\)

Bring all terms to one side:

\(x^2+(k-5)x+(k-2)=0.\)

For tangency, the discriminant is zero:

\((k-5)^2-4(k-2)=0.\)

Expand and simplify:

\(k^2-10k+25-4k+8=0.\)

So

\(k^2-14k+33=0.\)

Factorise:

\((k-11)(k-3)=0.\)

Hence

\(k=11 \quad\text{or}\quad k=3.\)

(ii) If \(k=11\), the tangent is

\(y=-5x+16\)

and the curve is

\(y=7-11x-x^2.\)

Equate them:

\(-5x+16=7-11x-x^2.\)

This gives

\(x^2+6x+9=0,\)

so

\((x+3)^2=0, \qquad x=-3.\)

Then

\(y=-5(-3)+16=31.\)

So the point of contact is \((-3,31)\).

If \(k=3\), the tangent is

\(y=-5x+8\)

and the curve is

\(y=7-3x-x^2.\)

Equate them:

\(-5x+8=7-3x-x^2.\)

This gives

\(x^2-2x+1=0.\)

So

\((x-1)^2=0, \qquad x=1.\)

Then

\(y=-5(1)+8=3.\)

So the point of contact is \((1,3)\).

(iii) The two points are \((-3,31)\) and \((1,3)\). Their distance is

\(\sqrt{(1-(-3))^2+(3-31)^2}.\)

So

\(d=\sqrt{4^2+(-28)^2} = \sqrt{16+784} = \sqrt{800} = 20\sqrt2.\)