Parallel plane containing a line

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Homework Statement
Find the plane that contains the line x=3+2T, Y=t, Z=8-t and is parallel to the plane 2x + 4y +8z =17.

The attempt at a solution

It seemed to me that since the plane is parallel to the plane 2x+4y+8z=17, it must be of the form 2x+4y+8z+D=0 so that it would be parallel.
(Is that true?)

Anyways, I then converted the equation of the line out of parametric into (x-3)/2 = Y = -(Z-8).
(I know that part's right.)

But now I'm not sure what to do. Ideas?
 
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Yes, your parallel plane equation is correct. Now just substitute x, y and z from the line equation into the plane equation and figure out what D must be.
 
It is, in fact, sufficient to use a single point on the line. Do you see why? For example, it should be easy to find D such that (3, 0, 8) (taking T= 0 in the equations of the line) is on the plane.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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