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Q.The equation of the plane parallel to the plane 3x−5y+4z=113x - 5y + 4z = 11 is

(a) 3x−5y+4z=213x - 5y + 4z = 21
(b) 3x+5y+4z=253x + 5y + 4z = 25
(c) 3x+5y+4z=353x + 5y + 4z = 35
(d) none of these
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Parallel planes share the same normal, so the coefficients of x,y,zx, y, z must be identical: 3x−5y+4z=213x - 5y + 4z = 21.

Two planes are parallel iff their normal vectors are proportional. The given plane 3x−5y+4z=113x - 5y + 4z = 11 has normal (3,−5,4)(3, -5, 4). A parallel plane must therefore have the form

3x−5y+4z=k3x - 5y + 4z = k …

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