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(A) Let the equation of the plane be $a(x - 1) + b(y - 1) + c(z - 1) = 0$. Since the plane passes through $(1, -1, 1)$,we have $a(1 - 1) + b(-1 - 1) +

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$3x - 4z + 1 = 0$

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(A) Let the equation of the plane be $a(x - 1) + b(y - 1) + c(z - 1) = 0$. Since the plane passes through $(1, -1, 1)$,we have $a(1 - 1) + b(-1 - 1) + c(1 - 1) = 0$,which implies $-2b = 0$,so $b = 0$. Now the equation becomes $a(x - 1) + c(z - 1) = 0$. Since it also passes through $(-7, -3, -5)$,we have $a(-7 - 1) + c(-5 - 1) = 0$,which gives $-8a - 6c = 0$,or $8a = -6c$,which simplifies to $a = -\frac{3}{4}c$. Substituting $a = -\frac{3}{4}c$ into the equation,we get $-\frac{3}{4}c(x - 1) + c(z - 1) = 0$. Dividing by $c$ (assuming $c 
eq 0$) and multiplying by $-4$,we get $3(x - 1) - 4(z - 1) = 0$,which simplifies to $3x - 3 - 4z + 4 = 0$,or $3x - 4z + 1 = 0$.

The equation of the plane passing through the three points $(1, 1, 1)$,$(1, -1, 1)$,and $(-7, -3, -5)$ is:

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