Vector Calculus....help please.....10 pts for best answer?

1) The vector connecting P to Q is (5,12,13) - (1,4,5) = (4, 8, 8).
So, PQ has equation (1, 4, 5) + t (4, 8, 8) for t in [0, 1].

Letting t = 2/3 yields the desired point:
(1, 4, 5) + (2/3) (4, 8, 8) = (11/3, 28/3, 31/3).
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2) I'll use s as the parameter (for minimal confusion) to the second line.

Equating like entries yields
1 - 2t = x = 0 + s ==> s = -2t
0 + 3t = y = 3 + 0s ==> t = 1
0 + 0t = z = 1 + s ==> s = -1.

So, the solution is t = 1 and s = -1 (which also satisfies the first relation).
Either parameter yields (x, y, z) = (-1, 3, 0).

I hope this helps!

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