1. Rewrite the equation using substitutions

Let $A=3x^2+4x+3,\qquad B=3x^2+4x+2.$ Then the given equation becomes $A^2-(k+1)AB+kB^2=0.$

Now factor this quadratic form: $A^2-(k+1)AB+kB^2=(A-B)(A-kB).$ This is because $(A-B)(A-kB)=A^2-kAB-AB+kB^2=A^2-(k+1)AB+kB^2.$

So the equation reduces to $(A-B)(A-kB)=0.$

Hence real roots exist if at least one of the following has a real solution:

1. $A-B=0$
2. $A-kB=0$

2. Solve the first factor

Since $A-B=(3x^2+4x+3)-(3x^2+4x+2)=1,$ we get $A-B=1\neq 0.$ So the first factor can never be zero.

Therefore we only need $A-kB=0.$

That is, $3x^2+4x+3-k(3x^2+4x+2)=0.$ Simplify: $(1-k)(3x^2+4x)+(3-2k)=0,$ so $3(1-k)x^2+4(1-k)x+(3-2k)=0.$

We need this equation to have real roots.

3. Case 1: $k=1$

Then the equation becomes $3-2(1)=1=0,$ which is impossible. So $k=1$ is not allowed.

4. Case 2: $k\neq 1$

Then it is a quadratic in $x$: $3(1-k)x^2+4(1-k)x+(3-2k)=0.$ For real roots, discriminant must satisfy $\Delta\ge 0.$

Compute: $\Delta=[4(1-k)]^2-4\cdot 3(1-k)(3-2k).$ So $\Delta=16(1-k)^2-12(1-k)(3-2k).$ Factor out $4(1-k)$: $\Delta=4(1-k)\big(4(1-k)-3(3-2k)\big).$ Now simplify inside: $4(1-k)-3(3-2k)=4-4k-9+6k=2k-5.$ Hence $\Delta=4(1-k)(2k-5).$

Thus real roots exist when $4(1-k)(2k-5)\ge 0,$ which is equivalent to $(1-k)(2k-5)\ge 0.$

5. Solve the inequality

Critical points are $k=1,\quad k=\frac52.$ Check intervals:

• For $k<1$: $(1-k)>0$, $(2k-5)<0$ so product $<0$.
• For $1<k<\frac52$: $(1-k)<0$, $(2k-5)<0$ so product $>0$.
• For $k>\frac52$: $(1-k)<0$, $(2k-5)>0$ so product $<0$.

Also, equality holds at $k=\frac52$. At $k=1$, discriminant is $0$, but we already separately found that the equation becomes impossible, so $k=1$ must be excluded.

Therefore, $k\in \left(1,\frac52\right].$

Given also $k>-1$, this does not change the set.

6. Match with options

The set is $\boxed{\left(1,\frac52\right]}$ which is Option A.

7. Comparison with stored answer

Stored correct answer: A

Our derived answer also gives A, so they agree.