A plane passing through the points A(2, 3, 5) | vedclass

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(C) Let the equation of the plane be $a(x - 2) + b(y - 3) + c(z - 5) = 0$. Since the plane passes through $(-3, -5, -7)$,we have $a(-3 - 2) + b(-5 - 3

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$(1, 4, 4)$

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(C) Let the equation of the plane be $a(x - 2) + b(y - 3) + c(z - 5) = 0$. Since the plane passes through $(-3, -5, -7)$,we have $a(-3 - 2) + b(-5 - 3) + c(-7 - 5) = 0$,which simplifies to $-5a - 8b - 12c = 0$,or $5a + 8b + 12c = 0$. The plane is perpendicular to $x - y + z = 1$,so the normal vector $(a, b, c)$ is perpendicular to $(1, -1, 1)$. Thus,$a - b + c = 0$,which means $a = b - c$. Substituting $a = b - c$ into $5a + 8b + 12c = 0$,we get $5(b - c) + 8b + 12c = 0$,so $13b + 7c = 0$. Let $b = 7$,then $c = -13$ and $a = 7 - (-13) = 20$. The equation of the plane is $20(x - 2) + 7(y - 3) - 13(z - 5) = 0$,which simplifies to $20x - 40 + 7y - 21 - 13z + 65 = 0$,or $20x + 7y - 13z + 4 = 0$. Checking the options: For $(1, 4, 4)$,$20(1) + 7(4) - 13(4) + 4 = 20 + 28 - 52 + 4 = 0$. Thus,the point $(1, 4, 4)$ lies on the plane.

$A$ plane passing through the points $A(2, 3, 5)$ and $B(-3, -5, -7)$ is perpendicular to the plane $x - y + z = 1$. Which of the following points lies on this plane?

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