Let a vector $$\vec{a}$$ has magnitude 9. Let a vector $$\vec{b}$$ be such that for every $$(x, y) \in \mathbb{R} \times \mathbb{R} - \{(0,0)\}$$, the vector $$x\vec{a} + y\vec{b}$$ is perpendicular to the vector $$6y\vec{a} - 18x\vec{b}$$. Then the value of $$|\vec{a} \times \vec{b}|$$ is equal to

We are given $$|\vec{a}| = 9$$ and for every $$(x, y) \neq (0, 0)$$, the vector $$(x\vec{a} + y\vec{b})$$ is perpendicular to $$(6y\vec{a} - 18x\vec{b})$$.

Since perpendicular vectors have zero dot product, we have $$(x\vec{a} + y\vec{b}) \cdot (6y\vec{a} - 18x\vec{b}) = 0$$. Expanding the dot product gives $$6xy|\vec{a}|^2 - 18x^2(\vec{a} \cdot \vec{b}) + 6y^2(\vec{a} \cdot \vec{b}) - 18xy|\vec{b}|^2 = 0$$, which can be grouped as $$6xy(|\vec{a}|^2 - 3|\vec{b}|^2) + 6(y^2 - 3x^2)(\vec{a} \cdot \vec{b}) = 0$$.

Because this identity holds for all $$(x, y) \neq (0,0)$$, the coefficients of the monomials must each vanish. Substituting $$x = 1, y = 0$$ into the grouped expression yields $$6(0 - 3)(\vec{a} \cdot \vec{b}) = 0$$, so $$\vec{a} \cdot \vec{b} = 0$$. Similarly, substituting $$x = 1, y = 1$$ gives $$6(|\vec{a}|^2 - 3|\vec{b}|^2) = 0$$, hence $$|\vec{a}|^2 = 3|\vec{b}|^2$$.

Using the given value of $$|\vec{a}|$$, we have $$81 = 3|\vec{b}|^2$$, which implies $$|\vec{b}|^2 = 27$$ and therefore $$|\vec{b}| = 3\sqrt{3}$$.

Since $$\vec{a} \cdot \vec{b} = 0$$, the vectors are perpendicular (θ = 90°). It follows that $$|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin(90°) = 9 \times 3\sqrt{3} \times 1 = 27\sqrt{3}$$.

The answer is Option B.