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Question

The determinant ∣ ∣xp+yxyyp+zyz0xp+yyp+z∣ ∣=0, if

A
x, y, z are in AP
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B
x, y, z are in GP
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C
x, y, z are in HP
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D
xy, yz, zx are in AP
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Solution

The correct option is B x, y, z are in GP
Given, ∣ ∣xp+yxyyp+zyz0xp+yyp+z∣ ∣=0
Applying C1C1(p C2+C3)
∣ ∣ ∣0xy0yz(xp2+yp+yp+z)xp+yyp+z∣ ∣ ∣=0 (xp2+2yp+z)(xzy2)=0 Either xp2+2yp+x=0 or y2=xz
x, y, z are in GP

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