$${(3{x^2} + 4x + 3)^2} - (k + 1)(3{x^2} + 4x + 3)(3{x^2} + 4x + 2) + k{(3{x^2} + 4x + 2)^2} = 0$$

Let $$3{x^2} + 4x + 3 = a$$

and $$3{x^2} + 4x + 2 = b \Rightarrow b = a - 1$$

Given equation becomes

$$ \Rightarrow {a^2} - (k + 1)ab + k{b^2} = 0$$

$$ \Rightarrow a(a - kb) - b(a - kb) = 0$$

$$ \Rightarrow (a - kb)(a - b) = 0 \Rightarrow a = kb$$ or a = b (reject) $$\because$$ a = kb

$$ \Rightarrow 3{x^2} + 4x + 3 = k(3{x^2} + 4x + 2)$$

$$ \Rightarrow 3(k - 1){x^2} + 4(k - 1)x + (2k - 3) = 0$$ for real roots

D $$\ge$$ 0

$$ \Rightarrow 16{(k - 1)^2} - 4(3(k - 1))(2k - 3) \ge 0$$

$$ \Rightarrow 4(k - 1)\{ 4(k - 1) - 3(2k - 3)\} \ge 0$$

$$ \Rightarrow 4(k - 1)\{ - 2k + 5\} \ge 0$$

$$ \Rightarrow - 4(k - 1)\{ 2k - 5\} \ge 0$$

$$ \Rightarrow (k - 1)(2k - 5) \le 0$$



$$\therefore$$ $$k \in \left[ {1,{5 \over 2}} \right]$$

$$\therefore$$ $$k \ne 1$$

$$\therefore$$ $$k \in \left( {1,{5 \over 2}} \right]$$