Let the area of the triangle formed by the lines $$x + 2 = y - 1 = z$$, $$\frac{x-3}{5} = \frac{y}{-1} = \frac{z-1}{1}$$ and $$\frac{x}{-3} = \frac{y-3}{3} = \frac{z-2}{1}$$ be A. Then $$A^2$$ is equal to _____.

To find the area $$A$$ of the triangle, we first need to determine its vertices by finding the points of intersection between the three given lines.

Given three lines are:

Line 1 ($$L_1$$): $$x + 2 = y - 1 = z$$

Line 2 ($$L_2$$): $$\frac{x-3}{5} = \frac{y}{-1} = \frac{z-1}{1}$$

Line 3 ($$L_3$$): $$\frac{x}{-3} = \frac{y-3}{3} = \frac{z-2}{1}$$

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Step 1: Finding Intersection of $$L_1$$ and $$L_2$$. Let it be A 

From $$L_1$$, we can express $$x$$ and $$y$$ in terms of $$z$$:

$$x = z - 2$$

$$y = z + 1$$

Substituting these into the equation of $$L_2$$:

$$\frac{(z - 2) - 3}{5} = \frac{z + 1}{-1} = \frac{z - 1}{1}$$

Comparing the last two parts:

$$\frac{z + 1}{-1} = z - 1 \implies -z - 1 = z - 1 \implies 2z = 0 \implies z = 0$$

Substituting $$z = 0$$ back gives:

$$x = 0 - 2 = -2$$

$$y = 0 + 1 = 1$$

Thus, the first vertex is $$A(-2, 1, 0)$$.

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Step 2: Finding Intersection of $$L_1$$ and $$L_3$$. Let it be B

Using the same expressions from $$L_1$$ ($$x = z - 2$$ and $$y = z + 1$$), substitute them into the equation of $$L_3$$:

$$\frac{z - 2}{-3} = \frac{(z + 1) - 3}{3} = \frac{z - 2}{1}$$

Comparing the first and third parts:

$$\frac{z - 2}{-3} = \frac{z - 2}{1} \implies z - 2 = -3(z - 2) \implies 4(z - 2) = 0 \implies z = 2$$

Substituting $$z = 2$$ back gives:

$$x = 2 - 2 = 0$$

$$y = 2 + 1 = 3$$

Thus, the second vertex is $$B(0, 3, 2)$$.

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Step 3: Finding Intersection of $$L_2$$ and $$L_3$$. Let it be C

Let a general point on $$L_2$$ be parameterized by $$t$$:

$$x = 5t + 3, \quad y = -t, \quad z = t + 1$$

Substitute these coordinates into the equation of $$L_3$$:

$$\frac{5t + 3}{-3} = \frac{-t - 3}{3} = \frac{(t + 1) - 2}{1}$$

$$\frac{5t + 3}{-3} = \frac{-t - 3}{3} = t - 1$$

Comparing the second and third parts:

$$\frac{-t - 3}{3} = t - 1 \implies -t - 3 = 3t - 3 \implies 4t = 0 \implies t = 0$$

Substituting $$t = 0$$ into the parametric coordinates gives:

$$x = 3, \quad y = 0, \quad z = 1$$

Thus, the third vertex is $$C(3, 0, 1)$$.

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Step 4: Calculate the Area $$A$$ using Vectors

Now form two adjacent side vectors from vertex $$A$$:

$$\vec{AB} = (0 - (-2))\hat{i} + (3 - 1)\hat{j} + (2 - 0)\hat{k} = 2\hat{i} + 2\hat{j} + 2\hat{k}$$

$$\vec{AC} = (3 - (-2))\hat{i} + (0 - 1)\hat{j} + (1 - 0)\hat{k} = 5\hat{i} - \hat{j} + \hat{k}$$

Find the cross product $$\vec{AB} \times \vec{AC}$$:

$$\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 2 & 2 \\ 5 & -1 & 1 \end{vmatrix}$$

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

$$= 4\hat{i} + 8\hat{j} - 12\hat{k}$$

The area $$A$$ of the triangle is half the magnitude of this cross product vector:

$$A = \frac{1}{2} \sqrt{4^2 + 8^2 + (-12)^2}$$

$$A = \frac{1}{2} \sqrt{16 + 64 + 144} = \frac{1}{2} \sqrt{224}$$

Squaring the area gives:

$$A^2 = \left(\frac{1}{2} \sqrt{224}\right)^2 = \frac{1}{4} \times 224 = 56$$