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

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(A) The equation of the first sphere is ${S_1} \equiv {x^2} + {y^2} + {z^2} + 6x - 8y - 2z - 13 = 0$. Its center ${C_1}$ is $(-3, 4, 1)$.The equation

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$-2$

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(A) The equation of the first sphere is ${S_1} \equiv {x^2} + {y^2} + {z^2} + 6x - 8y - 2z - 13 = 0$. Its center ${C_1}$ is $(-3, 4, 1)$.The equation of the second sphere is ${S_2} \equiv {x^2} + {y^2} + {z^2} - 10x + 4y - 2z - 8 = 0$. 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 the point $P(1, 1, 1)$,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 joining the centres 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 $a$ equals

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