Christmas Streak 57/88: 2018 CSAT (Korean SAT) #20 of Natural Sciences

Intuitive solution (the solution you should be using when taking the test)

If you want to maximize $d(\alpha),$ then $\alpha$ must be as distant as possible from all the three points.

This means that $\alpha$ is vertical to $\triangle \mathrm{ABC}$ and since $\alpha$ doesn't touch $\overline{\mathrm{AB}},$ we infer that $\alpha$ passes through the midpoint of $\overline{\mathrm{AC}}$ and $\overline{\mathrm{BC}}.$

Therefore, all of them are true.

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Formal solution (the solution you should be using when you're proving your answer to your teacher)

Let's say that $\alpha$ is not perpendicular to $\triangle \mathrm{ABC}.$

Then let $l$ be the line that $\alpha$ intersects with $\triangle \mathrm{ABC},$ and let $\gamma$ be the plane that is perpendicular to $\triangle \mathrm{ABC}$ and contains $l.$

The distances from the three points to $\alpha$ are always smaller than those to $l,$ (and consequently to $\gamma$ ) and therefore $d(\alpha)<d(\gamma).$

So $\beta\neq\alpha,$ and this means $\beta$ is not equal to any plane that is not perpendicular to $\triangle \mathrm{ABC}.$

Therefore $\beta$ is perpendicular to $\triangle \mathrm{ABC}.$ ( A. true)

Plane $\gamma$ is perpendicular to $\triangle \mathrm{ABC},$ and as said before, let $l$ be the intersection of $\gamma$ and $\triangle \mathrm{ABC}.$

Then saying that $\gamma$ is equivalent to $l$ on plane $\triangle \mathrm{ABC}$ does not lose generality.

Then let $\mathrm{M}$ be the midpoint of $\overline{\mathrm{AB}}$ and let $\delta$ be the plane that is perpendicular to $\triangle \mathrm{ABC},$ parallel to $\overline{\mathrm{AB}},$ and contains the intersection of $\overline{\mathrm{CM}}$ and $l,$ and let $m$ be the intersection of $\delta$ and $\triangle \mathrm{ABC}.$

Let's say that $l$ is not parallel to $\overline{\mathrm{AB}}.$

Then if the distance between $\mathrm{A}$ and $l,$ compared to the distance between $\mathrm{B}$ and $l,$ is:

$i)$ shorter, then $d(\gamma)$ is one of the distances from $\mathrm{A},~\mathrm{C}$ to $l.$ But since the distances from $\mathrm{A},~\mathrm{C}$ to $m$ is longer than each of those respectively, $d(\gamma)<d(\delta).$

$ii)$ longer, by similar reasons, $d(\gamma)<d(\delta).$

Then we know that $\beta\neq\gamma,$ and this means $\beta$ is not equal to any plane that is not parallel to $\overline{\mathrm{AB}}.$

Therefore $\beta$ is parallel to $\overline{\mathrm{AB}},$ and for it to be maximized, we need it to pass through the midpoint of $\overline{\mathrm{CM}},$ and consequently, it passes through both of midpoints of $\overline{\mathrm{AC}}$ and $\overline{\mathrm{BC}}.$ ( B. true)

It's trivial that the distances from $\mathrm{A},~\mathrm{B},~\mathrm{C}$ to $\beta$ are all equal, and can be proved by congruence. ( C. true)

From above, all of them are true.