If the plane 2ax - 3ay + 4az + 6 = 0 passes t | vedclass

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(A) The equation of the first sphere is $x^2 + y^2 + z^2 + 6x - 8y - 2z - 13 = 0$. Its center $C_1$ is $(-3, 4, 1)$.The equation of the second sphere

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$-2$

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(A) The equation of the first sphere is $x^2 + y^2 + z^2 + 6x - 8y - 2z - 13 = 0$. Its center $C_1$ is $(-3, 4, 1)$.The equation of the second sphere is $x^2 + y^2 + z^2 - 10x + 4y - 2z - 8 = 0$. Its center $C_2$ is $(5, -2, 1)$.The midpoint $P$ of the line segment joining $C_1$ and $C_2$ is given by $P = \left( \frac{-3 + 5}{2}, \frac{4 - 2}{2}, \frac{1 + 1}{2} \right) = (1, 1, 1)$.Since the plane $2ax - 3ay + 4az + 6 = 0$ passes through $P(1, 1, 1)$,we substitute the coordinates of $P$ into the plane equation:$2a(1) - 3a(1) + 4a(1) + 6 = 0$$2a - 3a + 4a + 6 = 0$$3a + 6 = 0$$3a = -6$$a = -2$.

If the plane $2ax - 3ay + 4az + 6 = 0$ passes through the midpoint of the line segment joining the centers of the spheres $x^2 + y^2 + z^2 + 6x - 8y - 2z = 13$ and $x^2 + y^2 + z^2 - 10x + 4y - 2z = 8$,then $a = ......$

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