If planes x - c y - b z = 0,c x - y + a z = 0 | vedclass

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(C) Since the given planes pass through a straight line,the determinant of the coefficients of the planes must be zero.$\begin{vmatrix} 1 & -c & -b \

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$1 - 2 a b c$

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(C) Since the given planes pass through a straight line,the determinant of the coefficients of the planes must be zero.$\begin{vmatrix} 1 & -c & -b \ c & -1 & a \ b & a & -1 \end{vmatrix} = 0$Expanding the determinant along the first row:$1( (-1)(-1) - (a)(a) ) - (-c)( (c)(-1) - (a)(b) ) + (-b)( (c)(a) - (-1)(b) ) = 0$$1(1 - a^2) + c(-c - ab) - b(ac + b) = 0$$1 - a^2 - c^2 - abc - abc - b^2 = 0$$1 - a^2 - b^2 - c^2 - 2abc = 0$Therefore,$a^2 + b^2 + c^2 = 1 - 2abc$.

If planes $x - c y - b z = 0$,$c x - y + a z = 0$ and $b x + a y - z = 0$ pass through a straight line then $a^2 + b^2 + c^2 =$

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