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(D) The general term in the multinomial expansion is given by:$T_{r_1, r_2, r_3} = \frac{10!}{r_1! r_2! r_3!} (3x^3)^{r_1} (-2x^2)^{r_2} (5x^{-5})^{r_

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$9$

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(D) The general term in the multinomial expansion is given by:$T_{r_1, r_2, r_3} = \frac{10!}{r_1! r_2! r_3!} (3x^3)^{r_1} (-2x^2)^{r_2} (5x^{-5})^{r_3}$$T_{r_1, r_2, r_3} = \frac{10!}{r_1! r_2! r_3!} (3)^{r_1} (-2)^{r_2} (5)^{r_3} x^{3r_1 + 2r_2 - 5r_3}$For the constant term,the power of $x$ must be $0$:$3r_1 + 2r_2 - 5r_3 = 0$  $(1)$Also,$r_1 + r_2 + r_3 = 10$  $(2)$From $(2)$,$r_2 = 10 - r_1 - r_3$. Substituting into $(1)$:$3r_1 + 2(10 - r_1 - r_3) - 5r_3 = 0$$3r_1 + 20 - 2r_1 - 2r_3 - 5r_3 = 0$$r_1 + 20 = 7r_3$Testing integer values for $r_3$:If $r_3 = 3$,then $r_1 = 7(3) - 20 = 1$. Then $r_2 = 10 - 1 - 3 = 6$.Constant term $= \frac{10!}{1! 6! 3!} (3)^1 (-2)^6 (5)^3$$= \frac{10 	imes 9 	imes 8 	imes 7}{3 	imes 2 	imes 1} 	imes 3 	imes 64 	imes 125$$= (10 	imes 3 	imes 4 	imes 7) 	imes 3 	imes 2^6 	imes 5^3$$= (840) 	imes 3 	imes 2^6 	imes 125 = (2^3 	imes 3 	imes 5 	imes 7) 	imes 3 	imes 2^6 	imes 5^3$$= 2^9 	imes 3^2 	imes 5^4 	imes 7^1 = 2^9 \cdot l$,where $l = 3^2 \cdot 5^4 \cdot 7^1$ is an odd integer.Thus,$k = 9$.

If the constant term in the expansion of $\left(3 x^{3}-2 x^{2}+\frac{5}{x^{5}}\right)^{10}$ is $2^{k} \cdot l$,where $l$ is an odd integer,then the value of $k$ is equal to

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