Let A(5, -3),B(3, -2),and C(-1, 5) be three p | vedclass

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(D) Let the coordinates of point $P$ be $(x, y)$.Given the condition $PA^2 + 2PB^2 = 3PC^2$.Substituting the coordinates of $A(5, -3)$,$B(3, -2)$,and

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$\left(2, \frac{37}{22}\right)$

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(D) Let the coordinates of point $P$ be $(x, y)$.Given the condition $PA^2 + 2PB^2 = 3PC^2$.Substituting the coordinates of $A(5, -3)$,$B(3, -2)$,and $C(-1, 5)$:$(x - 5)^2 + (y + 3)^2 + 2[(x - 3)^2 + (y + 2)^2] = 3[(x + 1)^2 + (y - 5)^2]$Expanding the terms:$(x^2 - 10x + 25 + y^2 + 6y + 9) + 2(x^2 - 6x + 9 + y^2 + 4y + 4) = 3(x^2 + 2x + 1 + y^2 - 10y + 25)$$(x^2 + y^2 - 10x + 6y + 34) + 2(x^2 + y^2 - 6x + 4y + 13) = 3(x^2 + y^2 + 2x - 10y + 26)$$3x^2 + 3y^2 - 22x + 14y + 60 = 3x^2 + 3y^2 + 6x - 30y + 78$$-22x - 6x + 14y + 30y + 60 - 78 = 0$$-28x + 44y - 18 = 0$Dividing by $-2$:$14x - 22y + 9 = 0$Checking option $D$ $\left(2, \frac{37}{22}\right)$:$14(2) - 22\left(\frac{37}{22}\right) + 9 = 28 - 37 + 9 = 0$.Thus,the point $\left(2, \frac{37}{22}\right)$ lies on the locus.

Let $A(5, -3)$,$B(3, -2)$,and $C(-1, 5)$ be three points. If $P$ is a point satisfying the condition $PA^2 + 2PB^2 = 3PC^2$,then a point that lies on the locus of $P$ is

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