Write the required line through $(-1,2,1)$

The line $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{4}$ has direction ratios $(2,3,4)$.

So the line passing through $(-1,2,1)$ and parallel to it is $\frac{x+1}{2}=\frac{y-2}{3}=\frac{z-1}{4}=\lambda$

Hence its parametric form is $x=-1+2\lambda,\quad y=2+3\lambda,\quad z=1+4\lambda.$

Write the second line in parametric form

Given $\frac{x+2}{3}=\frac{y-3}{2}=\frac{z-4}{1}=\mu$

so $x=-2+3\mu,\quad y=3+2\mu,\quad z=4+\mu.$

Find their point of intersection $P$

At intersection, $-1+2\lambda=-2+3\mu \quad ...(1)$ $2+3\lambda=3+2\mu \quad ...(2)$ $1+4\lambda=4+\mu \quad ...(3)$

From (1): $2\lambda-3\mu=-1$

From (2): $3\lambda-2\mu=1$

Solve these two equations:

Multiply first by $3$ and second by $2$: $6\lambda-9\mu=-3$ $6\lambda-4\mu=2$

Subtract: $5\mu=5 \Rightarrow \mu=1$

Then from $3\lambda-2\mu=1$: $3\lambda-2=1 \Rightarrow 3\lambda=3 \Rightarrow \lambda=1.$

Check in (3): $1+4(1)=5,\quad 4+1=5$ so it is consistent.

Therefore, $P=(-1+2,\;2+3,\;1+4)=(1,5,5).$

Find distance from $P$ to $Q(4,-5,1)$

Using distance formula: $PQ=\sqrt{(4-1)^2+(-5-5)^2+(1-5)^2}$ $=\sqrt{3^2+(-10)^2+(-4)^2}$ $=\sqrt{9+100+16}$ $=\sqrt{125}=5\sqrt{5}.$

Match with options

$5\sqrt{5}$ corresponds to Option C.

Compare with stored correct answer

Stored correct answer: C

Our derived answer: C

So they agree.