A plane pi is passing through the points A(1, | vedclass

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(A) Let the equation of the plane $\pi$ be $a(x - 1) + b(y + 2) + c(z - 3) = 0$,where $\vec{n} = (a, b, c)$ is the normal vector to the plane. Since t

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$\frac{63}{\sqrt{594}}$

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(A) Let the equation of the plane $\pi$ be $a(x - 1) + b(y + 2) + c(z - 3) = 0$,where $\vec{n} = (a, b, c)$ is the normal vector to the plane. Since the plane passes through $B(6, 4, 5)$,we have $a(6 - 1) + b(4 + 2) + c(5 - 3) = 0$,which simplifies to $5a + 6b + 2c = 0$. The plane $\pi$ is perpendicular to the plane $3x - y + z = 2$,whose normal vector is $\vec{n_1} = (3, -1, 1)$. Thus,the dot product of the normal vectors is zero: $3a - b + c = 0$. Solving the system of equations $5a + 6b + 2c = 0$ and $3a - b + c = 0$: From the second equation,$b = 3a + c$. Substituting into the first: $5a + 6(3a + c) + 2c = 0 \implies 23a + 8c = 0$. Let $a = 8$,then $c = -23$. Then $b = 3(8) - 23 = 24 - 23 = 1$. So,the normal vector is $\vec{n} = (8, 1, -23)$. The equation of the plane $\pi$ is $8(x - 1) + 1(y + 2) - 23(z - 3) = 0$,which simplifies to $8x + y - 23z + 63 = 0$. The perpendicular distance from $(0, 0, 0)$ to the plane is $d = \frac{|8(0) + 1(0) - 23(0) + 63|}{\sqrt{8^2 + 1^2 + (-23)^2}} = \frac{63}{\sqrt{64 + 1 + 529}} = \frac{63}{\sqrt{594}}$.

$A$ plane $\pi$ is passing through the points $A(1, -2, 3)$ and $B(6, 4, 5)$. If the plane $\pi$ is perpendicular to the plane $3x - y + z = 2$,then the perpendicular distance from $(0, 0, 0)$ to the plane $\pi$ is

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