If two distinct point $$Q$$, $$R$$ lie on the line of intersection of the planes $$-x + 2y - z = 0$$ and $$3x - 5y + 2z = 0$$ and $$PQ = PR = \sqrt{18}$$ where the point $$P$$ is $$(1, -2, 3)$$, then the area of the triangle $$PQR$$ is equal to

We find the line of intersection of planes $$-x + 2y - z = 0$$ and $$3x - 5y + 2z = 0$$.

The direction of the line of intersection is given by the cross product of the normals:

$$\vec{n_1} = (-1, 2, -1), \quad \vec{n_2} = (3, -5, 2)$$

$$\vec{d} = \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 2 & -1 \\ 3 & -5 & 2 \end{vmatrix}$$

$$= \hat{i}(4 - 5) - \hat{j}(-2 + 3) + \hat{k}(5 - 6) = (-1, -1, -1)$$

Direction vector: $$(1, 1, 1)$$.

Both planes pass through the origin, so the line passes through $$(0, 0, 0)$$. The parametric form is:

$$(x, y, z) = (t, t, t)$$

Point $$P = (1, -2, 3)$$. The foot of perpendicular from $$P$$ to the line is:

$$\vec{PM} = (t-1, t+2, t-3)$$, which must be perpendicular to $$(1,1,1)$$:

$$(t-1) + (t+2) + (t-3) = 0 \implies 3t - 2 = 0 \implies t = \frac{2}{3}$$

Foot of perpendicular $$M = \left(\frac{2}{3}, \frac{2}{3}, \frac{2}{3}\right)$$.

$$PM^2 = \left(\frac{1}{3}\right)^2 + \left(\frac{8}{3}\right)^2 + \left(\frac{7}{3}\right)^2 = \frac{1 + 64 + 49}{9} = \frac{114}{9} = \frac{38}{3}$$

Since $$PQ = PR = \sqrt{18}$$, and $$Q, R$$ lie on the line, triangle $$PQR$$ is isosceles with $$PQ = PR$$. The distance from $$P$$ to the line is $$PM = \sqrt{\frac{38}{3}}$$.

In triangle $$PMQ$$: $$MQ^2 = PQ^2 - PM^2 = 18 - \frac{38}{3} = \frac{54 - 38}{3} = \frac{16}{3}$$

$$MQ = \frac{4}{\sqrt{3}}$$

Since $$M$$ is the midpoint of $$QR$$ (by symmetry of the isosceles triangle), $$QR = 2 \cdot MQ = \frac{8}{\sqrt{3}}$$.

Area of triangle $$PQR$$:

$$= \frac{1}{2} \times QR \times PM = \frac{1}{2} \times \frac{8}{\sqrt{3}} \times \sqrt{\frac{38}{3}}$$

$$= \frac{4}{\sqrt{3}} \times \frac{\sqrt{38}}{\sqrt{3}} = \frac{4\sqrt{38}}{3}$$

Hence the correct answer is Option B: $$\dfrac{4}{3}\sqrt{38}$$.