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(B) The equation of a plane passing through the point $(2, 2, 1)$ is given by $a(x - 2) + b(y - 2) + c(z - 1) = 0$. Since the plane passes through $(9

Vedclass · Accepted Answer

$3x + 4y - 5z = 9$

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(B) The equation of a plane passing through the point $(2, 2, 1)$ is given by $a(x - 2) + b(y - 2) + c(z - 1) = 0$. Since the plane passes through $(9, 3, 6)$,we have $a(9 - 2) + b(3 - 2) + c(6 - 1) = 0$,which simplifies to $7a + b + 5c = 0$. Since the plane is perpendicular to $2x + 6y + 6z - 1 = 0$,the normal vectors are perpendicular,so $2a + 6b + 6c = 0$,which simplifies to $a + 3b + 3c = 0$. Using the cross product of the direction vectors $(7, 1, 5)$ and $(1, 3, 3)$ to find the normal vector $(a, b, c)$: $a = (1 	imes 3) - (5 	imes 3) = 3 - 15 = -12$ $b = (5 	imes 1) - (7 	imes 3) = 5 - 21 = -16$ $c = (7 	imes 3) - (1 	imes 1) = 21 - 1 = 20$ Dividing by $-4$,we get the direction ratios $(a, b, c) = (3, 4, -5)$. Substituting these into the plane equation: $3(x - 2) + 4(y - 2) - 5(z - 1) = 0$. $3x - 6 + 4y - 8 - 5z + 5 = 0 \Rightarrow 3x + 4y - 5z = 9$.

Find the equation of the plane passing through the points $(2, 2, 1)$ and $(9, 3, 6)$ and perpendicular to the plane $2x + 6y + 6z - 1 = 0$.

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