1. We are given:

• $|\vec a| = 9$
• For every $(x,y) \neq (0,0)$, $x\vec a + y\vec b \perp 6y\vec a - 18x\vec b$

This means their dot product is zero for all real $x,y$: $(x\vec a + y\vec b) \cdot (6y\vec a - 18x\vec b)=0$

2. Expand the dot product:

$x\vec a \cdot 6y\vec a + x\vec a \cdot (-18x\vec b) + y\vec b \cdot 6y\vec a + y\vec b \cdot (-18x\vec b)=0$

Using commutativity of dot product, $6xy|\vec a|^2 - 18x^2(\vec a\cdot \vec b) + 6y^2(\vec a\cdot \vec b) - 18xy|\vec b|^2 = 0$

So, $6xy|\vec a|^2 - 18xy|\vec b|^2 + ( -18x^2 + 6y^2)(\vec a\cdot\vec b)=0$

Since this holds for all $x,y$, the coefficients of $x^2$, $y^2$, and $xy$ must separately be zero.

3. Compare coefficients:

• Coefficient of $x^2$: $-18(\vec a\cdot\vec b)=0 \implies \vec a\cdot\vec b=0$

• Coefficient of $y^2$: $6(\vec a\cdot\vec b)=0$ which is consistent.

• Coefficient of $xy$: $6|\vec a|^2 - 18|\vec b|^2 = 0$ $|\vec a|^2 = 3|\vec b|^2$

Given $|\vec a|=9$, $81 = 3|\vec b|^2$ $|\vec b|^2=27$ $|\vec b|=3\sqrt{3}$

4. Now compute $|\vec a \times \vec b|$.

We know $|\vec a \times \vec b| = |\vec a|\,|\vec b|\sin\theta$

Since $\vec a\cdot\vec b=0$, the angle between them is $90^\circ$, so $\sin\theta=1$.

Hence, $|\vec a \times \vec b| = 9\cdot 3\sqrt{3} = 27\sqrt{3}$

5. Therefore the correct option is: $\boxed{27\sqrt{3}}$ which is option B.