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 C1→C1−(pC2+C3) ⇒∣∣
∣
∣∣0xy0yz−(xp2+yp+yp+z)xp+yyp+z∣∣
∣
∣∣=0⇒−(xp2+2yp+z)(xz−y2)=0∴Eitherxp2+2yp+x=0ory2=xz ⇒ x, y, z are in GP