Solving the following quadratic equations by factorization method:

(i) $x^{2}+10 i x-21=0$

$\Rightarrow x^{2}+7 i x+3 i x-21=0$

$\Rightarrow x(x+7 i)+3 i(x+7 i)=0$

$\Rightarrow(x+7 i)(x+3 i)=0$

$\Rightarrow(x+7 i)=0$ or $(x+3 i)=0$

$\Rightarrow x=-7 i,-3 i$

So, the roots of the given quadratic equation are $-3 i$ and $-7 i$.

(ii) $x^{2}+(1-2 i) x-2 i=0$

$\Rightarrow x^{2}+x-2 i x-2 i=0$

$\Rightarrow x(x+1)-2 i(x+1)=0$

$\Rightarrow(x+1)(x-2 i)=0$

$\Rightarrow(x+1)=0$ or $(x-2 i)=0$

$\Rightarrow x=-1,2 i$

So, the roots of the given quadratic equation are $-1$ and $2 i$.

(iii) $x^{2}-(2 \sqrt{3}+3 i) x+6 \sqrt{3} i=0$

$\Rightarrow x^{2}-2 \sqrt{3} x-3 i x+6 \sqrt{3} i=0$

$\Rightarrow x(x-2 \sqrt{3})-3 i(x-2 \sqrt{3})=0$

$\Rightarrow(x-2 \sqrt{3})(x-3 i)=0$

$\Rightarrow(x-2 \sqrt{3})=0$ or $(x-3 i)=0$

$\Rightarrow x=2 \sqrt{3}, 3 i$

So, the roots of the given quadratic equation are $2 \sqrt{3}$ and $3 i$.

(iv) $6 x^{2}-17 i x-12=0$

$\Rightarrow 6 x^{2}-9 i x-8 i x-12=0$

$\Rightarrow 3 x(2 x-3 i)-4 i(2 x-3 i)=0$

$\Rightarrow(2 x-3 i)(3 x-4 i)=0$

$\Rightarrow(2 x-3 i)=0$ or $(3 x-4 i)=0$

$\Rightarrow x=\frac{3}{2} i, \frac{4}{3} i$

So, the roots of the given quadratic equation are $\frac{3}{2} i$ and $\frac{4}{3} i$.