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

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Step 1: Rewrite the equation of the first line in standard form.

The given equation is $\frac{x}{2}=\frac{2y-6}{4}=\frac{1-z}{-1}$. We can rewrite this as:\r\n\r\n$\frac{x}{2}=\frac{y-3}{2}=\frac{z-1}{1}$

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Step 2: Identify the direction vector of the line.

From the standard form, the direction vector $\\vec{b}$ of the line is $\\vec{b} = <2, 2, 1>$.

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Step 3: Find a point on the first line.

Let's find a point on the first line. By setting the parameter to 0, we get the point $A(0, 3, 1)$.

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Step 4: Identify the point on the second line.

The second line passes through the point $B(4, 0, -5)$.

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Step 5: Find the vector connecting the two points.

The vector $\\vec{AB}$ connecting the points $A$ and $B$ is given by:\r\n\r\n$\\vec{AB} = <4-0, 0-3, -5-1> = <4, -3, -6>$

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Step 6: Calculate the cross product of the direction vector and the vector connecting the points.

We need to find the cross product of $\\vec{b}$ and $\\vec{AB}$:\r\n\r\n$\\vec{b} \\times \\vec{AB} = \\begin{vmatrix} \\hat{i} & \\hat{j} & \\hat{k} \\\\ 2 & 2 & 1 \\\\ 4 & -3 & -6 \\end{vmatrix} = \\hat{i}(-12+3) - \\hat{j}(-12-4) + \\hat{k}(-6-8) = -9\\hat{i} + 16\\hat{j} - 14\\hat{k}$\r\n\r\nSo, $\\vec{b} \\times \\vec{AB} = <-9, 16, -14>$

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Step 7: Find the magnitude of the cross product.

The magnitude of $\\vec{b} \\times \\vec{AB}$ is:\r\n\r\n$||\\vec{b} \\times \\vec{AB}|| = \\sqrt{(-9)^2 + (16)^2 + (-14)^2} = \\sqrt{81 + 256 + 196} = \\sqrt{533}$

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Step 8: Find the magnitude of the direction vector.

The magnitude of $\\vec{b}$ is:\r\n\r\n$||\\vec{b}|| = \\sqrt{2^2 + 2^2 + 1^2} = \\sqrt{4 + 4 + 1} = \\sqrt{9} = 3$

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Step 9: Calculate the distance between the lines.

The distance $d$ between the two lines is given by:\r\n\r\n$d = \\frac{||\\vec{b} \\times \\vec{AB}||}{||\\vec{b}||} = \\frac{\\sqrt{533}}{3}$

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\r\n Final Answer: $\\frac{\\sqrt{533}}{3}$<\/span>\r\n <\/p>\r\n <\/div>\r\n <\/div>

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