Let $$P$$ be the plane passing through the intersection of the planes $$\vec{r} \cdot (\hat{i} + 3\hat{j} - \hat{k}) = 5$$ and $$\vec{r} \cdot (2\hat{i} - \hat{j} + \hat{k}) = 3$$, and the point $$(2, 1, -2)$$. Let the position vectors of the points $$X$$ and $$Y$$ be $$\hat{i} - 2\hat{j} + 4\hat{k}$$ and $$5\hat{i} - \hat{j} + 2\hat{k}$$ respectively. Then the points

We need to find the plane $$P$$ passing through the intersection of $$\vec{r} \cdot (\hat{i} + 3\hat{j} - \hat{k}) = 5$$ and $$\vec{r} \cdot (2\hat{i} - \hat{j} + \hat{k}) = 3$$, and through the point $$(2, 1, -2)$$.

Since it passes through the intersection of the given planes, its equation can be written as $$(x + 3y - z - 5) + \lambda(2x - y + z - 3) = 0$$ which expands to $$(1 + 2\lambda)x + (3 - \lambda)y + (-1 + \lambda)z = 5 + 3\lambda$$.

Substituting the point $$(2, 1, -2)$$ into this family yields $$(1 + 2\lambda)(2) + (3 - \lambda)(1) + (-1 + \lambda)(-2) = 5 + 3\lambda$$ which simplifies to $$2 + 4\lambda + 3 - \lambda + 2 - 2\lambda = 5 + 3\lambda$$ and hence $$7 + \lambda = 5 + 3\lambda \implies \lambda = 1$$.

Therefore the required plane has equation $$3x + 2y + 0 \cdot z = 8$$, i.e., $$3x + 2y = 8$$.

Defining $$f(x,y) = 3x + 2y - 8$$ allows us to check the sides of various points; points on the same side have the same sign of $$f$$. For $$X = (1, -2, 4)$$ we have $$f = 3(1) + 2(-2) - 8 = -9 < 0$$ whereas for $$Y = (5, -1, 2)$$ we obtain $$f = 3(5) + 2(-1) - 8 = 5 > 0$$.
Similarly, for $$X + Y = (6, -3, 6)$$ one finds $$f = 3(6) + 2(-3) - 8 = 4 > 0$$; for $$Y - X = (4, 1, -2)$$, $$f = 3(4) + 2(1) - 8 = 6 > 0$$; and for $$X - Y = (-4, -1, 2)$$, $$f = 3(-4) + 2(-1) - 8 = -22 < 0$$.

Since $$X$$ yields a negative value of $$f$$ and $$Y$$ yields a positive value, they lie on opposite sides of the plane. Hence the answer is Option C: $$X$$ and $$Y$$ are on opposite sides of $$P$$.