Find the equation of the plane passing through the points $(2, 1, -1)$ and $(-1, 3, 4)$ and perpendicular to the plane $x - 2y + 4z = 10$.
(D) Let the equation of the plane be $a(x - 2) + b(y - 1) + c(z + 1) = 0$ ... $(i)$
Since it passes through $(-1, 3, 4)$,we have $a(-1 - 2) + b(3 - 1) + c(4 + 1) = 0$,which simplifies to $-3a + 2b + 5c = 0$ ... $(ii)$
The plane $(i)$ is perpendicular to $x - 2y + 4z = 10$,so the normal vectors are perpendicular: $1(a) - 2(b) + 4(c) = 0$,which gives $a - 2b + 4c = 0$ ... $(iii)$
Solving $(ii)$ and $(iii)$ using cross multiplication:
$\frac{a}{(2)(4) - (5)(-2)} = \frac{-b}{(-3)(4) - (5)(1)} = \frac{c}{(-3)(-2) - (2)(1)}$
$\frac{a}{8 + 10} = \frac{-b}{-12 - 5} = \frac{c}{6 - 2}$
$\frac{a}{18} = \frac{b}{17} = \frac{c}{4} = k$
Thus,$a = 18k, b = 17k, c = 4k$.
Substituting these into $(i)$: $18k(x - 2) + 17k(y - 1) + 4k(z + 1) = 0$
$18x - 36 + 17y - 17 + 4z + 4 = 0$
$18x + 17y + 4z - 49 = 0$
Therefore,the equation is $18x + 17y + 4z = 49$.
Since it passes through $(-1, 3, 4)$,we have $a(-1 - 2) + b(3 - 1) + c(4 + 1) = 0$,which simplifies to $-3a + 2b + 5c = 0$ ... $(ii)$
The plane $(i)$ is perpendicular to $x - 2y + 4z = 10$,so the normal vectors are perpendicular: $1(a) - 2(b) + 4(c) = 0$,which gives $a - 2b + 4c = 0$ ... $(iii)$
Solving $(ii)$ and $(iii)$ using cross multiplication:
$\frac{a}{(2)(4) - (5)(-2)} = \frac{-b}{(-3)(4) - (5)(1)} = \frac{c}{(-3)(-2) - (2)(1)}$
$\frac{a}{8 + 10} = \frac{-b}{-12 - 5} = \frac{c}{6 - 2}$
$\frac{a}{18} = \frac{b}{17} = \frac{c}{4} = k$
Thus,$a = 18k, b = 17k, c = 4k$.
Substituting these into $(i)$: $18k(x - 2) + 17k(y - 1) + 4k(z + 1) = 0$
$18x - 36 + 17y - 17 + 4z + 4 = 0$
$18x + 17y + 4z - 49 = 0$
Therefore,the equation is $18x + 17y + 4z = 49$.
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