Given line and its direction vector

The line $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{-2}$ has direction ratios $\vec d=(2,3,-2).$

Required line

We need a line passing through the point $(0,1,2)$ and perpendicular to the given line.

So if the required line has direction ratios $(l,m,n)$, then they must satisfy $2l+3m-2n=0.$

Check each option

Since all options pass through $(0,1,2)$, we only need to test their direction ratios.

Option A: $\frac{x}{3}=\frac{y-1}{-4}=\frac{z-2}{3}$ Direction ratios: $(3,-4,3)$

Dot product with $(2,3,-2)$: $2(3)+3(-4)-2(3)=6-12-6=-12\neq 0$ So, A is not perpendicular.

Option B: $\frac{x}{3}=\frac{y-1}{4}=\frac{z-2}{-3}$ Direction ratios: $(3,4,-3)$

Dot product: $2(3)+3(4)-2(-3)=6+12+6=24\neq 0$ So, B is not perpendicular.

Option C: $\frac{x}{-3}=\frac{y-1}{4}=\frac{z-2}{3}$ Direction ratios: $(-3,4,3)$

Dot product: $2(-3)+3(4)-2(3)=-6+12-6=0$ So, C is perpendicular.

Option D: $\frac{x}{3}=\frac{y-1}{4}=\frac{z-2}{3}$ Direction ratios: $(3,4,3)$

Dot product: $2(3)+3(4)-2(3)=6+12-6=12\neq 0$ So, D is not perpendicular.

Conclusion

The only line through $(0,1,2)$ perpendicular to the given line is $\boxed{\frac{x}{-3}=\frac{y-1}{4}=\frac{z-2}{3}}$ which is Option C.