Find the equation of the plane passing through the point (– 1,

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 Multiple Choice QuestionsShort Answer Type

191. In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them:
2x + y + 3z – 2 = 0 and x – 2y + 5 = 0
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192. In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them:
2x – 2y + 4z + 5 = 0 and 3x – 3y + 6z – 1 = 0
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193. In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them:
2x – y + 3z – 1 = 0 and 2x – y + 3z + 3 = 0
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194. In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them:
4x + 8y + z – 8 = 0 and y + z – 4 = 0
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195. Find the equation of the plane which bisects the line joining the points (–1, 2, 3) and (3, – 5 6) at right angles.
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196. Find the equation of the plane through the intersection of the planes x + y + z = 9 and 2 x + 3 y + 4 z + 5 = 0 and passing through the point (1, 1, 1).
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 Multiple Choice QuestionsLong Answer Type

197. Find the equation of plane passing through origin and intersection of planes 2x – 3y + z = 9, x – y + z = 4 
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 Multiple Choice QuestionsShort Answer Type

198. Find the equation of the plane through the intersection of the planes 3x – y + 2 z – 4 = 0 and x + y + z – 2 = 0 and the point (2, 2, 1).
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199. Find the equation of the plane passing through the point (– 1, – 1, 2) and perpendicular to each of the following planes:
2x + 3y – 3 = 2 and 5x – 4y + z = 6.


The equation of any plane through (– 1,–1, 2) is
a(x + 1) + b(y + 1) + c(z – 2) = 0    ...(1)
∴ it is perpendicular to the planes
2x + 3y – 3z = 2 and 5x – 4y + z = 6
∴ 2a + 3b – 3c = 0    ...(2)
and 5a – 4b + c = 0    ...(3)
Solving (2) and (3), we get,
                    fraction numerator straight a over denominator 3 minus 12 end fraction space equals space fraction numerator straight b over denominator negative 15 minus 2 end fraction space equals space fraction numerator straight c over denominator negative 8 minus 15 end fraction

therefore space space space space space space space space space space space space space fraction numerator straight a over denominator negative 9 end fraction space equals space fraction numerator straight b over denominator negative 17 end fraction equals space fraction numerator straight c over denominator negative 23 end fraction
therefore space space space space space space space space space space space space space space space space straight a over 9 space equals space straight b over 17 space equals space straight c over 28 space equals space straight k comma space space say.
therefore space space space straight a space equals space 9 space straight k comma space space space 6 space equals space 17 space straight k comma space space space straight c space equals space 23 space straight k

∴ a = 9 k, 6 = 17 k, c = 23 k
Putting values of a, b, c in (1), we get
9 k (x + 1)+ 17 k (y + 1) + 23 k (z – 2) = 0
or 9 (x + 1) + 17 (+ 1) + 23 (z – 2) = 0
or 9 x + 9 + 17 y + 17 + 23 z – 46 = 0
or 9 x + 17 y + 23 z – 20 = 0
which is required equation of plane.

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 Multiple Choice QuestionsLong Answer Type

200. Find the direction ratios of the normal to the plane passing through the point (2,1, 3) and the line of intersection of the planes x + 2 y + z = 3 and 2 x – y – z = 5.
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