Let the vertices Q and R of the triangle PQR lie on the line $$\frac{x+3}{5} = \frac{y-1}{2} = \frac{z+4}{3}$$, $$QR = 5$$ and the coordinates of the point P be $$(0, 2, 3)$$. If the area of the triangle PQR is $$\frac{m}{n}$$, then:

The line having both Q and R can be written in parametric form as
$$(x,y,z)=\left(5t-3,\;2t+1,\;3t-4\right),\quad t\in\mathbb{R}\; -(1)$$

Let $$t=a$$ give point $$Q$$ and $$t=b$$ give point $$R$$:
$$Q\;(5a-3,\;2a+1,\;3a-4),\qquad R\;(5b-3,\;2b+1,\;3b-4)$$

Set $$k=b-a$$. Then the vector $$\overrightarrow{QR}$$ is
$$\overrightarrow{QR}=(5k,\;2k,\;3k)$$

Its length is given to be $$5$$:
$$|\overrightarrow{QR}|=\lvert k\rvert\sqrt{5^{2}+2^{2}+3^{2}} =\lvert k\rvert\sqrt{38}=5$$
$$\Rightarrow\;\lvert k\rvert=\dfrac{5}{\sqrt{38}}\; -(2)$$

Point $$P$$ is fixed at $$(0,2,3)$$. Write the vectors from $$P$$ to $$Q$$ and $$R$$:
$$\overrightarrow{PQ}=Q-P=(5a-3,\;2a-1,\;3a-7)$$
$$\overrightarrow{PR}=R-P=\overrightarrow{PQ}+\overrightarrow{QR}$$

The area of $$\triangle PQR$$ is
$$\text{Area}=\dfrac12\, \lvert\overrightarrow{PQ}\times\overrightarrow{PR}\rvert$$

Because $$\overrightarrow{PR}=\overrightarrow{PQ}+\overrightarrow{QR}$$, and $$\overrightarrow{PQ}\times\overrightarrow{PQ}=0$$, we have
$$\overrightarrow{PQ}\times\overrightarrow{PR} =\overrightarrow{PQ}\times\overrightarrow{QR}$$

Compute the cross-product. Put
$$\overrightarrow{PQ}=(5a-3,\;2a-1,\;3a-7),\; \overrightarrow{QR}=(5k,\;2k,\;3k)$$

$$$ \begin{aligned} \overrightarrow{PQ}\times\overrightarrow{QR} &=\begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\ 5a-3 & 2a-1 & 3a-7\\ 5k & 2k & 3k \end{vmatrix}\\[4pt] &=k\bigl(11,\,-26,\,-1\bigr) \end{aligned} $$$

$$\lvert\overrightarrow{PQ}\times\overrightarrow{QR}\rvert =|k|\sqrt{11^{2}+(-26)^{2}+(-1)^{2}} =|k|\sqrt{121+676+1}=|k|\sqrt{798}$$

Using $$-(2)$$, $$|k|=\dfrac{5}{\sqrt{38}}$$, hence
$$\text{Area}= \dfrac12\cdot\dfrac{5}{\sqrt{38}}\cdot\sqrt{798} =\dfrac{5}{2}\sqrt{\dfrac{798}{38}} =\dfrac{5}{2}\sqrt{21}\; -(3)$$

Thus $$\text{Area}=\dfrac{5\sqrt{21}}{2}$$. If the area is written as $$\dfrac{m}{n}$$, comparison with $$-(3)$$ gives
$$\dfrac{m}{n}=\dfrac{5\sqrt{21}}{2} \;\Longrightarrow\;2m-5\sqrt{21}\,n=0$$

Therefore the relation between $$m$$ and $$n$$ is
Option B: $$2m-5\sqrt{21}n=0$$