### Video Transcript

Find the parametric form of the
equation of the plane that passes through the points π΄ one, five, one; π΅ three,
four, three; and πΆ two, three, four.

Okay, so we have this plane that
contains these three points π΄, π΅, and πΆ. And we want to figure out which of
these five options gives us the parametric form of the equation of the plane. We can recall that to define the
parametric form of a planeβs equation, we need to know one point on the plane and
two vectors that lie in it. Here we have three points. And we can actually use these
points to define two coplanar vectors. For example, if we subtract point
π΄ from point π΅, we get this vector in pink. And likewise, if we subtract π΄
from πΆ, we get this vector.

Substituting in the coordinates of
these points, for π΅ minus π΄, three minus one is two, four minus five is negative
one, and three minus one is two. And then, for πΆ minus π΄, two
minus one is one, three minus five is negative two, and four minus one is three. And what weβre saying is that these
coordinates are actually the components of two vectors in our plane. Weβll call them π― one and π―
two.

And now note that we have a point β
weβve picked point π΄ β as well as two vectors that lie in this plane. We can now recall the most general
way to write the parametric form of a planeβs equation. Itβs in terms of a point in the
plane that we have a vector going to and two vectors, π― one and π― two, that lie in
the plane, each of which is multiplied by its own parameter.

Applying this equation to our
scenario, we can write that the vector π« with components π₯, π¦, π§ is equal to a
vector to our known point, one, five, one, plus a parameter π‘ one times our first
vector π― one plus a second parameter π‘ two times π― two. Note that from this expression, we
can get equations for π₯ and π¦ and π§. For example, π₯ is equal to one
plus two times π‘ one plus one times π‘ two. π¦ is equal to five minus one times
π‘ one minus two times π‘ two. And then π§ equals one plus two
times π‘ one plus three times π‘ two. And these equations altogether are
the parametric form of the equation of our plane. Looking over our answer options, we
see a match with choice (E).

Now, itβs true that there are many
equivalent ways to express the parametric form of a planeβs equation. For example, we mightβve picked a
different point, say π΅ or πΆ, rather than point π΄. And we mightβve solved for
different coplanar vectors than these two here. Nonetheless, we still wouldnβt have
chosen options (A), (B), (C), or (D) because none of those options use point π΄, π΅,
or πΆ as the point that lies in the plane.

Our final answer then is choice
(E). π₯ equals one plus two times π‘ one
plus π‘ two, π¦ equals five minus π‘ one minus two π‘ two, and π§ equals one plus
two π‘ one plus three π‘ two.