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(D) The equation of the line passing through $(-1,2,1)$ and parallel to the line $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{4}$ is given by:$\frac{x+1}{2}=

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$5 \sqrt{5}$

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(D) The equation of the line passing through $(-1,2,1)$ and parallel to the line $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{4}$ is given by:$\frac{x+1}{2}=\frac{y-2}{3}=\frac{z-1}{4}=\lambda$So,the coordinates of any point on this line are $(2\lambda-1, 3\lambda+2, 4\lambda+1)$.The equation of the second line is given by:$\frac{x+2}{3}=\frac{y-3}{2}=\frac{z-4}{1}=\mu$So,the coordinates of any point on this line are $(3\mu-2, 2\mu+3, \mu+4)$.For the intersection point $P$,the coordinates must be equal:$2\lambda-1 = 3\mu-2 \implies 2\lambda - 3\mu = -1$  $(i)$$3\lambda+2 = 2\mu+3 \implies 3\lambda - 2\mu = 1$  (ii)Solving $(i)$ and (ii),we multiply $(i)$ by $2$ and (ii) by $3$:$4\lambda - 6\mu = -2$$9\lambda - 6\mu = 3$Subtracting these gives $5\lambda = 5$,so $\lambda = 1$.Substituting $\lambda = 1$ into $(i)$,$2(1) - 3\mu = -1 \implies 3\mu = 3 \implies \mu = 1$.Checking with the $z$-coordinate: $4(1)+1 = 5$ and $1+4 = 5$. Since they match,the intersection point $P$ is $(2(1)-1, 3(1)+2, 4(1)+1) = (1, 5, 5)$.The distance of $P(1, 5, 5)$ from $Q(4, -5, 1)$ is:$PQ = \sqrt{(4-1)^2 + (-5-5)^2 + (1-5)^2}$$PQ = \sqrt{3^2 + (-10)^2 + (-4)^2}$$PQ = \sqrt{9 + 100 + 16} = \sqrt{125} = 5\sqrt{5}$.

Let the line passing through the points $(-1,2,1)$ and parallel to the line $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{4}$ intersect the line $\frac{x+2}{3}=\frac{y-3}{2}=\frac{z-4}{1}$ at the point $P$. Then the distance of $P$ from the point $Q(4,-5,1)$ is:

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