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The answer is yes. Tahmid can always choose such 3 points.

We will prove this using double counting. We will count the number of isosceles triangle.

Defining:

Number the vertices 1 through 17.

Call an isosceles triangle with all different colored vertices a $trichromatic$ triangle. Assume that there is $n$ of them.

Call an isosceles triangle with 2 same-colored vertices a $dichromatic$ triangle.

Call an isosceles triangle with 3 same-colored vertices a $monochromatic$ triangle. Assume that there is $m$ of them.

For an isosceles triangle, call its common point of the equal sides its $center$

Call the difference of the assigned numbers of its center with the number at another vertice, its $power$. Of course, the power is in modulo 17.

Let $R,B,Y$ be the number of Red, Blue, Yellow points respectively.

First Counting: 

In a isosceles triangle, $its$ power can be anything from $[1,8]$ and its center can be chosen from the 17 points.

So, There are total $17\cdot 8 = 136$ isosceles triangle.

No triangle has been counted twice because there is no equilateral triangle.

Now, for the sake of contradiction, assume that all isosceles triangles have at least 2 points of the same color.

Second counting:

One segment is a part of exactly 3 isosceles triangle. One with the extra vertice as the center and the rest with the endpoints of the given segment as the center.

Again, these three triangles can not be the same because there is no equilateral triangle.

If we count based on the edges with same colored endpoints, over-count occurs in case of a monochromatic triangle, they are counted thrice.

The number of edges with both red colored endpoints is $\binom{R}{2}$. Similar for blue and yellow.

So, the number of isosceles triangle is $3(\binom{R}{2}+\binom{B}{2}+\binom{Y}{2}) - 2m + n$

Combining the two counting,

$136 = 3(\binom{R}{2}+\binom{B}{2}+\binom{Y}{2}) - 2m + n$

For the sake of contradiction, assume that $n=0$.

So, $136 = \frac{3}{2}(R^2-R+B^2-B+Y^2-Y)-2m$

$\implies 2(68 + m) = \frac{3}{2}(R^2+B^2+Y^2-17)$

$\implies 2\mid \frac{3}{2}(R^2+B^2+Y^2-17)$

$\implies 4\mid (R^2+B^2+Y^2-17)$

$\implies R^2+B^2+Y^2 \equiv 1 \pmod{4}$

But, $R,B,G$ are odd. So, $R^2\equiv B^2\equiv Y^2\equiv1\pmod4$

So, $R^2+B^2+Y^2 \equiv 3 \pmod{4}$.

Contradiction.

So, Tahmid can always find an isosceles triangle with different colored points.