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Let =|A xx^2 1B y y^2 1C z z^2 1|a n d1=...

Let `=|A xx^2 1B y y^2 1C z z^2 1|a n d_1=|A B C x y z y z z xx y|,` then show that `_1=`

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If a x1 2+b y1 2+c z1 2=a x1 2+b y2 2+c z2 2=a x3 2+b y3 2+c z3 2=d\ a n d\ a x_2x_3+b y_2y_3+c z_2z_3=a x_3x_1+b y_3y_1+c z_3z_1=a x_1x_2+b y_1y_2+c z_1z_2=f then prove that |x_1y_1z_1x_2y_2z_1x_3y_3z_3|=(d-f)[(d+2f)/(a b c)]^(//2)(a , b ,\ c!=0)

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If a x_1^2+b y_1^2+c z_1^2=a x_2 ^2+b y_2 ^2+c z_2 ^2=a x_3 ^2+b y_3 ^2+c z_3 ^2=d ,a x_2 x_3+b y_2y_3+c z_2z_3=a x_3x_1+b y_3y_1+c z_3z_1=a x_1x_2+b y_1y_2+c z_1z_2=f, then prove that |(x_1, y_1, z_1), (x_2, y_2, z_2), (x_3,y_3,z_3)|=(d-f){((d+2f))/(a b c)}^(1//2)

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