The equation of the given line is:

$(2,7,-3)t + (2,1,5)$

This can be rewritten as the parametric equations of the line: $x = 2t + 2, \quad y = 7t + 1, \quad z = -3t + 5$

The direction vector $\mathbf{v}_1$ of the line is $(2,7,-3)$.

To find the equation of a line perpendicular to this given line that passes through the point $(5,2,8)$, we need to use the fact that the direction vector $\mathbf{v}_2$ of the new line must be perpendicular to $\mathbf{v}_1$.

So, let $\mathbf{v}_2 = (a, b, c)$ be the direction vector of the new line. Since the two lines are perpendicular, their direction vectors must satisfy the dot product condition:

$\mathbf{v}_1 \cdot \mathbf{v}_2 = 0$

Substituting the direction vector of the given line:

$(2,7,-3) \cdot (a,b,c) = 2a + 7b - 3c = 0$

This is the equation for the direction vector of the new line. Now, the parametric form of the new line passing through the point $(5,2,8)$ can be written as:

$(x, y, z) = (5, 2, 8) + t(a, b, c)$

To summarize:

1. Solve $2a + 7b - 3c = 0$ to find $(a,b,c)$, the direction vector of the new line.
2. Substitute into the parametric equation to get the full form of the line.

Would you like to continue with specific values or further details?

Here are 5 related questions for further understanding:

1. How is the dot product used to find perpendicularity between vectors?
2. What are parametric equations of a line?
3. How do you find the equation of a line passing through a point?
4. Can we calculate the angle between two lines in space?
5. How do you use cross product to find a perpendicular vector?

Tip: The dot product between two perpendicular vectors is always zero!