I don't think so!

Relevant wiki: Multinomial Theorem

$\text {From a Property of Binomial Thm.} \\ \text{Coefficient of a term in } (a+b+c+d+...)^n \text { is :} \\ \dfrac{n!}{{n_1!}\times{n_2!}\times{n_3!}\times{n_4!}\times{....}}\times{a^{n_1}}\times{b^{n_2}}\times{c^{n_3}}\times{d^{n_4}\times{....}} \\ \text{Where } n_1+n_2+n_3+n_4+\cdots=n \text{ and } 0\le{n_1,n_2,n_3,n_4,...}\le{n}\\ \text{In this case } \dfrac{20!}{ {n_1!}\times{n_2!}\times{n_3!}}\times{1^{n_1}}\times{x^{5n_2}}\times{x^{7n_3}} \\n_1+n_2+n_3=20 \text{ and } 2n_2+7n_3=17 \text{ (given)} \\ \text{Possible values of }n_2=2\text{ and }n_1=1 \\ \implies \dfrac{20!}{ {17!}\times{2!}\times{1!}}\times{1^{19}}\times{x^{10}}\times{x^7} \\ \implies \dfrac{20\times19\times18}{2}x^{10+7} \\ \text{co-efficient of }x^{17} = \color{#3D99F6}{\boxed{3420}}$

We're looking for the coefficient of x^17 in (1 + x^5 + x^7)^20.

The only way to make x^17 from 1, x^5, and x^7 is (x^5)(x^5)(x^7), so out of the 20 (1 + x^5 + x^7)s, we have to pick 2 to get the x^5 from, and 1 to get the x^7 from. 20c2 = 190. 18 choices for the x^7 makes the total combinations 190*18 = x^17 will be made 3420 different ways, making the coefficient of x^17 3420 .