Detailed Solution<\/h3>\r\n <\/div>\r\n\r\n

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Step 1: Find the direction vector of AB

The direction vector $\\vec{AB}$ is given by the difference of the position vectors of points B and A.

$\\vec{AB} = (1 - (-1), -2 - 2, 5 - 1) = (2, -4, 4)$

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Step 2: Find a vector normal to both AB and the line CD

The direction vector of the line CD is $\\vec{d} = (1, -2, 2)$. The vector normal to both $\\vec{AB}$ and $\\vec{d}$ is given by their cross product.

$\\vec{n} = \\vec{AB} \\times \\vec{d} = \\begin{vmatrix} \\hat{i} & \\hat{j} & \\hat{k} \\\\ 2 & -4 & 4 \\\\ 1 & -2 & 2 \\end{vmatrix} = \\hat{i}(-8 - (-8)) - \\hat{j}(4 - 4) + \\hat{k}(-4 - (-4)) = (0, 0, 0)$

Since the cross product is the zero vector, $\\vec{AB}$ and $\\vec{d}$ are parallel. This means that AB and CD are parallel, as expected in a parallelogram.

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Step 3: Find the distance between the parallel lines AB and CD

Let's take a point on the line CD, say $P(4, -7, 8)$. The distance between the point A and the line CD is the same as the distance between the lines AB and CD.

The vector $\\vec{AP} = (4 - (-1), -7 - 2, 8 - 1) = (5, -9, 7)$

The distance $d$ between the point A and the line CD is given by

$d = \\frac{|\\vec{AP} \\times \\vec{d}|}{|\\vec{d}|}$

$\\vec{AP} \\times \\vec{d} = \\begin{vmatrix} \\hat{i} & \\hat{j} & \\hat{k} \\\\ 5 & -9 & 7 \\\\ 1 & -2 & 2 \\end{vmatrix} = \\hat{i}(-18 - (-14)) - \\hat{j}(10 - 7) + \\hat{k}(-10 - (-9)) = (-4, -3, -1)$

$|\\vec{AP} \\times \\vec{d}| = \\sqrt{(-4)^2 + (-3)^2 + (-1)^2} = \\sqrt{16 + 9 + 1} = \\sqrt{26}$

$|\\vec{d}| = \\sqrt{1^2 + (-2)^2 + 2^2} = \\sqrt{1 + 4 + 4} = \\sqrt{9} = 3$

Therefore, the distance $d = \\frac{\\sqrt{26}}{3}$

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Step 4: Find the length of side AB

$|\\vec{AB}| = \\sqrt{2^2 + (-4)^2 + 4^2} = \\sqrt{4 + 16 + 16} = \\sqrt{36} = 6$

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Step 5: Find the area of the parallelogram ABCD

The area of the parallelogram is given by the product of the base and the height. Here, the base is $|\\vec{AB}|$ and the height is the distance $d$ between the parallel lines.

Area = $|\\vec{AB}| \\times d = 6 \\times \\frac{\\sqrt{26}}{3} = 2\\sqrt{26}$

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\r\n Final Answer: Distance between AB and CD is $\\frac{\\sqrt{26}}{3}$ and Area of parallelogram ABCD is $2\\sqrt{26}$<\/span>\r\n <\/p>\r\n <\/div>\r\n <\/div>

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