A plane x passes through the point (1, 1, 1). | vedclass

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(C) First,we find the prime factorization of $2001$. $2001 = 3 	imes 667 = 3 	imes 23 	imes 29$. Given that $a, b, c$ are factors of $2001$ such th

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$23x + 29y + 3z = 55$

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(C) First,we find the prime factorization of $2001$. $2001 = 3 	imes 667 = 3 	imes 23 	imes 29$. Given that $a, b, c$ are factors of $2001$ such that $a The direction ratios of the normal to the plane are given as $b, c, a$,which are $23, 29, 3$. The equation of a plane with normal vector $(b, c, a)$ passing through $(x_0, y_0, z_0)$ is $b(x - x_0) + c(y - y_0) + a(z - z_0) = 0$. Substituting the values,we get $23(x - 1) + 29(y - 1) + 3(z - 1) = 0$. Expanding this,we get $23x - 23 + 29y - 29 + 3z - 3 = 0$. $23x + 29y + 3z - 55 = 0$. Therefore,the equation of the plane is $23x + 29y + 3z = 55$.

$A$ plane $x$ passes through the point $(1, 1, 1)$. If $b, c, a$ are the direction ratios of a normal to the plane,where $a, b, c$ $(a < b < c)$ are the factors of $2001$,then the equation of the plane is

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