<p>Well, let's go Algebra 2 way.</p>

<ol>

<li>x-coordinates of points A( 1,3,0) and C(-1,0,2) are opposite numbers.</li>

<li>Points A and C are equidistant from plane yz.</li>

<li>Midpoint M of AC is in plane yz.</li>

<li>Coordinates of M:

(1,3,0) + (-1,0,2) ) / 2 = (0,3/2,1).

See footnotes 4.1 and 4.2.</li>

<li>h is a height of triangle ABC from A to BM.</li>

<li>Area of triangle AMB:

A(AMB) = 1/2 (BM) h.</li>

<li>BM is median of triangle ABC.

A(AMB) = A(CMB),

A(ABC) = (2) A(AMB),

A(ABC) = (BM) h.</li>

<li>Distance between B(0,2,5) and M(0,3/2,1)

BM = sqrt(65) / 2 .</li>

<li>Since line BM is in yz-plane and goes through points B(2,5) and M(1/2,1),

its equation is z = (8)y - 11, or

(8)y + (-1)z + (-11) = 0.

(x=0 for all points of line BM).

Equation of line BM in xyz-space:

ax + by + cz + d = 0, or

(0)x + (8)y + (-1)z + (-11) = 0, </li>

<li>h is a distance between point A(1,3,0) and line BM.

Formula:

h = | a(1) + b(3) + c(0) + d | / sqrt(a^2 + b^2 + c^2)

h = | (0)(1) + (8)(3) + (-10)(0) + (-11) | / sqrt (0^2 + 8^2 + (-1)^2)

h = 13 / sqrt(65)</li>

<li>A(ABC) = (BM) h.

A(ABC) = (sqrt(65) / 2 ) * ( 13 / sqrt(65) ) = 13 / 2 = 6.5</li>

</ol>

<p>6.5 !!!!!!!!

And all the work :(

It's gotta be some much shorter solution.

This one was just the most direct that came to my head.</p>

<h2>I also hope I did not make some dumb mistakes.</h2>

<p>Footnotes.

4.1 Instead of wading through steps 1-3 we could just find coordinates of midpoint M and notice that x-coordinate of M(0,3/2,1) is 0.<br>

That means point M is in plane yz.

4.2 On the other hand, even step 1 might not be obvious.

I strongly suggest you carefully draw 3-D diagram of triangle ABC in xyz coordinate system.

If you look at AC projection into xy-plane (say, AD), and AD y-intercept (say, N), you may notice that AN=ND, so yz-plane "cuts" AC in half, and AC midpoint M is in yz-plane (M projection into xy-plane is N).

Sounds scary, but drawing this could turn to be a lot of fun.</p>