Q.

The four planes 7x + 4y – 4z + 2 = 0,  36x–51y + 12z + 17 = 0, 14x+8y–8z–12= 0 and 12x–17y+4z–1=0  are the four faces of a

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a

Parallelepiped

b

rectangular parallelepiped

c

Cube

d

Tetrahedron

answer is B.

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Detailed Solution

. 14x + 3y – 8 z + 4 = 0

14x + 8y – 8 z – 12 = 0

d1=4+12142+82+82

36x -  51y + 12z+12 = 0

36x – 51y +12z – 3 = 0

d2=17+3(36)2+(51)2+122

d1d2

The adjacent planes are perpendicular so that the planes are forms rectangular parallalopipe 

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