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Explanation:
Let P be the origin (0,0) and Q be the point where this parametric line intersects the xy plane. The goal is to find the distance from P to Q. First we need to locate Q.
Because Q is on the xy plane, this means it's of the form (x,y,0). We could say it's in the form (x,y), but writing (x,y,0) allows for the 3rd dimension to take account.
This means we'll plug z = 0 into the third equation to find t
z = 6t-12
0 = 6t-12
12 = 6t
6t = 12
t = 12/6
t = 2
Now we have enough info to pinpoint the exact location of point Q.
Point Q is located at (6,8) which is the same as (6,8,0)
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P is at (0,0) and Q is at (6,8)
Apply the distance formula to finish off the problem.
[tex]d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(0-6)^2 + (0-8)^2}\\\\d = \sqrt{(-6)^2 + (-8)^2}\\\\d = \sqrt{36 + 64}\\\\d = \sqrt{100}\\\\d = 10\\\\[/tex]
Alternatively, you can use the pythagorean theorem [tex]a^2+b^2 = c^2[/tex] to plug in a = 6 and b = 8 to find that c = 10.