### Video Transcript

In which of the following planes does the point three, negative one, five lie? (A) Three π₯ minus π¦ plus five π§ equals zero. (B) Two π₯ minus four π¦ plus π§ plus five equals zero. (C) Two π₯ plus π¦ minus two π§ plus 23 equals zero. (D) π₯ minus two π¦ plus two π§ minus 15 equals zero. And (E) negative four π₯ minus four π¦ plus two π§ plus seven equals zero.

Okay, so here we have these equations for five different planes, and we want to find in which one of these planes the point three, negative one, five lies. Weβll know that a certain plane does contain this point if when we plug in three for π₯, negative one for π¦, and five for π§, that gives a value of zero, the value on the right-hand side of each of these five equations. So, for example, considering first option (A), if we substitute in three for π₯, negative one for π¦, and five for π§, then when we calculate this left-hand side of that planeβs equation, we get nine plus one plus 25. This equals 35. And since this result is different from zero, it means that the point weβve picked, three, negative one, five, does not lie in this plane. So option (A) isnβt our answer.

But letβs move on to check option (B). Two times three minus four times negative one plus five plus five equals six plus four plus five plus five. And this adds up to 20. Since 20 also is not zero, the equation written in option (B) also isnβt our answer. Moving on to option (C), two times three plus negative one minus two times five plus 23 equal six minus one minus 10 plus 23, or 18, once again, not zero. Option (C) then is out of consideration as well. In option (D), we have three minus two times negative one plus two times five minus 15 or three plus two plus 10 minus 15. And this all does add up to zero. The plane in option (D) then does contain our point of interest.

Just to confirm that this is the only correct answer, letβs look lastly at option (E). Negative four times three minus four times negative one plus two times five plus seven equals negative 12 plus four plus 10 plus seven, which added together equals positive nine. The plane in option (E) then does not contain our point. So the plane containing the point three, negative one, five has the equation π₯ minus two π¦ plus two π§ minus 15 equals zero.