1. Identify the two circles

Given: $C: x^2+y^2=4$ So for $C$, center is $(0,0)$ and radius is $r_1=2.$

Also, $C': x^2+y^2-4\lambda x+9=0.$ Rewrite it: $x^2-4\lambda x+y^2+9=0$ $(x-2\lambda)^2+y^2=4\lambda^2-9.$ So for $C'$, center is $(2\lambda,0)$ and radius $r_2=\sqrt{4\lambda^2-9}.$ For this to be a real circle, we need $4\lambda^2-9>0 \implies |\lambda|>\frac32.$ (Though this will automatically follow from intersection condition.)

2. Condition for two distinct intersection points

Distance between centers: $d=|2\lambda|=2|\lambda|.$

For two circles to intersect at two distinct points: $|r_1-r_2|<d<r_1+r_2.$

That is, $|2-\sqrt{4\lambda^2-9}|<2|\lambda|<2+\sqrt{4\lambda^2-9}.$

We solve both inequalities.

3. Solve the right inequality

$2|\lambda|<2+\sqrt{4\lambda^2-9}.$ Since both sides are nonnegative, rearrange: $2|\lambda|-2<\sqrt{4\lambda^2-9}.$ Squaring is safe when $|\lambda|\ge 1$, and in our valid region $|\lambda|>\frac32$ anyway.

$\left(2|\lambda|-2\right)^2<4\lambda^2-9$ $4\lambda^2-8|\lambda|+4<4\lambda^2-9$ $-8|\lambda|+4<-9$ $8|\lambda|>13$ $|\lambda|>\frac{13}{8}.$

4. Solve the left inequality

$|2-\sqrt{4\lambda^2-9}|<2|\lambda|.$ This is equivalent to $-2|\lambda|<2-\sqrt{4\lambda^2-9}<2|\lambda|.$ The right part gives back the previous inequality type; the left part gives $\sqrt{4\lambda^2-9}<2+2|\lambda|,$ which is always true. So the effective condition remains $|\lambda|>\frac{13}{8}.$

Hence the set of all such values is $(-\infty,-13/8)\cup(13/8,\infty)=\mathbb R-\left[-\frac{13}{8},\frac{13}{8}\right].$ Therefore, $a=-\frac{13}{8},\qquad b=\frac{13}{8}.$

5. Compute the point

We need $(8a+12,\;16b-20).$ Substitute: $8a+12=8\left(-\frac{13}{8}\right)+12=-13+12=-1,$ $16b-20=16\left(\frac{13}{8}\right)-20=26-20=6.$ So the point is $(-1,6).$

6. Check which curve contains $(-1,6)$

Option A:

$x^2+2y^2-5x+6y=3$ Substitute $(-1,6)$: $1+2(36)-5(-1)+6(6)=1+72+5+36=114\ne 3.$ Not correct.

Option B:

$5x^2-y=-11$ $5(1)-6=-1\ne -11.$ Not correct.

Option C:

$x^2-4y^2=7$ $1-4(36)=1-144=-143\ne 7.$ Not correct.

Option D:

$6x^2+y^2=42$ $6(1)+36=42.$ Correct.

7. Final answer

The point lies on: $\boxed{6x^2+y^2=42}$ So the correct option is D.