If the constant term in the expansion of (3x^3-2x^2+ 5 x^5)^10 is 2^k.l, where l is an odd integer, then the value of k is equal to

For ${(3{x}^{3}-2{x}^{2}+\frac{5}{{x}^{5}})}^{10}$(3x3−2x2+x55​)10the general term is given by

$${T}_{r+1}=\frac{10!}{{r}_{1}!{r}_{2}!{r}_{3}!}{(3)}^{{r}_{1}}{(-2)}^{{r}_{2}}{(5)}^{{r}_{3}}{(x)}^{3{r}_{1}+2{r}_{2}-5{r}_{3}}$$Tr+1​=r1​!r2​!r3​!10!​(3)r1​(−2)r2​(5)r3​(x)3r1​+2r2​−5r3​

For term independent of $x$x the exponent $3{r}_{1}+2{r}_{2}-5{r}_{3}=0\ldots (1)$3r1​+2r2​−5r3​=0…(1)

Also we know ${r}_{1}+{r}_{2}+{r}_{3}=10\cdots (2)$r1​+r2​+r3​=10⋯(2)

From $(1)$(1) and $(2)$(2), we get

$${r}_{1}+2(10-{r}_{3})-5{r}_{3}=0$$r1​+2(10−r3​)−5r3​=0

i.e. ${r}_{1}+20=7{r}_{3}$r1​+20=7r3​

So ${r}_{1},{r}_{2},{r}_{3}=1,6,3$r1​,r2​,r3​=1,6,3

Hence the constant term $=\frac{10!}{1!6!3!}{(3)}^{1}{(-2)}^{6}{(5)}^{3}$=1!6!3!10!​(3)1(−2)6(5)3

$$={2}^{9}\cdot {3}^{2}\cdot {5}^{4}\cdot {7}^{1}$$=29⋅32⋅54⋅71

$$\Rightarrow k=9$$⇒k=9