1. Given vectors

$$\vec a=(1,2,3),\qquad \vec b=(1,1,-1)$$

Let

$$\vec c=(x,y,z).$$

We need to find $|\vec a\times \vec c|^2$.

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2. Use the condition $\vec a\cdot \vec c=11$

$$\vec a\cdot \vec c=x+2y+3z=11 \qquad ...(1)$$

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3. Use the condition $\vec b\cdot(\vec a\times \vec c)=27$

This is the scalar triple product:

$$\vec b\cdot(\vec a\times \vec c)=[\vec b\,\vec a\,\vec c].$$

First compute

$$\vec a\times \vec c= \begin{vmatrix} \hat i & \hat j & \hat k\\ 1&2&3\\ x&y&z \end{vmatrix} =(2z-3y)\hat i-(z-3x)\hat j+(y-2x)\hat k.$$

So

$$\vec b\cdot(\vec a\times \vec c) =(1,1,-1)\cdot(2z-3y,\,3x-z,\,y-2x).$$

Therefore,

$$(2z-3y)+(3x-z)-(y-2x)=27$$ $$5x-4y+z=27 \qquad ...(2)$$

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4. Use the condition $\vec b\cdot \vec c=-\sqrt3|\vec b|$

First,

$$|\vec b|=\sqrt{1^2+1^2+(-1)^2}=\sqrt3.$$

Hence

$$\vec b\cdot \vec c=-\sqrt3\cdot \sqrt3=-3.$$

So,

$$x+y-z=-3 \qquad ...(3)$$

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5. Solve the linear system

We have:

$$x+2y+3z=11 \quad ...(1)$$ $$5x-4y+z=27 \quad ...(2)$$ $$x+y-z=-3 \quad ...(3)$$

From (3):

$$x=-3-y+z.$$

Substitute into (1):

$$(-3-y+z)+2y+3z=11$$ $$y+4z=14 \qquad ...(4)$$

Substitute into (2):

$$5(-3-y+z)-4y+z=27$$ $$-15-9y+6z=27$$ $$-3y+2z=14 \qquad ...(5)$$

Now solve (4) and (5):

From (4),

$$y=14-4z.$$

Put into (5):

$$-3(14-4z)+2z=14$$ $$-42+12z+2z=14$$ $$14z=56$$ $$z=4.$$

Then

$$y=14-16=-2.$$

From (3):

$$x+(-2)-4=-3$$ $$x-6=-3$$ $$x=3.$$

Thus,

$$\vec c=(3,-2,4).$$

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6. Compute $\vec a\times \vec c$

$$\vec a\times \vec c= \begin{vmatrix} \hat i & \hat j & \hat k\\ 1&2&3\\ 3&-2&4 \end{vmatrix}$$ $$=\hat i(2\cdot 4-3\cdot(-2))- \hat j(1\cdot 4-3\cdot 3)+ \hat k(1\cdot(-2)-2\cdot 3)$$ $$=14\hat i+5\hat j-8\hat k.$$

Therefore,

$$|\vec a\times \vec c|^2=14^2+5^2+(-8)^2=196+25+64=285.$$

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7. Final answer

$$|\vec a\times \vec c|^2=285$$