Let the lengths of the sides $AB, \ BC$, and $CA$ of a triangle $ABC$ be $a, \ b$ and $c$, respectively.
Also, let the foot of the perpendicular line drawn from the vertex $A$ to $BC$ be $H$, and let $AH=h$.
Then, for the right-angled triangle $ABH$,
$$BH=\sqrt{AB^2-AH^2}=\sqrt{a^2-h^2},$$
Similarly, for the right-angled triangle $ACH$,
$$CH=\sqrt{AC^2-AH^2}=\sqrt{c^2-h^2},$$
Since $BC=BH+CH$,
$$BC=\sqrt{a^2-h^2}+\sqrt{c^2-h^2}=b,$$
$$∴ \sqrt{a^2-h^2}=b-\sqrt{c^2-h^2}. \qquad [1]$$
Squaring both sides of $[1]$,
$$a^2-h^2=b^2-2b \sqrt{c^2-h^2}+c^2-h^2,$$
$$∴ \ \sqrt{c^2-h^2}=\frac{b^2+c^2-a^2}{2b}. \qquad [2]$$

Squaring both sides of $[2]$,
$$c^2-h^2=\frac{a^4+b^4+c^4-2a^2 b^2-2a^2 c^2+2b^2 c^2}{4b^2},$$
$$∴ \ h^2=\frac{2a^2 b^2+2a^2 c^2+2b^2 c^2-a^4-b^4-c^4}{4b^2},$$
$$∴ \ h=\frac{\sqrt{2a^2 b^2+2a^2 c^2+2b^2 c^2-a^4-b^4-c^4}}{2b}.$$

If the area of ​​this triangle is $S$,
$$S=\frac{bh}{2}=\frac{\sqrt{2a^2 b^2+2a^2 c^2+2b^2 c^2-a^4-b^4-c^4}}{4}, \qquad [3]$$
Also, if the diameter of the inscribed circle is $d$,
$$S=\frac{1}{2}×\frac{d}{2}×(a+b+c)=\frac{(a+b+c)d}{4}. \qquad [4]$$
From $[3]$ and $[4]$,
$$\frac{(a+b+c)d}{4}=\frac{\sqrt{2a^2 b^2+2a^2 c^2+2b^2 c^2-a^4-b^4-c^4}}{4},$$
$$∴ \ d=\frac{\sqrt{2a^2 b^2+2a^2 c^2+2b^2 c^2-a^4-b^4-c^4}}{a+b+c}. \qquad [5]$$
Substituting $a=10, \ b=21$ and $c=17$ into $[5]$, we get
$$d=\frac{\sqrt{2×10^2×21^2+2×10^2×17^2+2×21^2×17^2-10^4-21^4-17^4}}{10+21+17}$$
$$=\frac{\sqrt{88200+57800+254898-10000-194481-83521}}{48}=\frac{336}{48}=7.$$