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

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

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

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(C) First,we find the prime factors of $2001$. $2001 = 3 	imes 23 	imes 29$. Given that $a, b, c$ are the factors 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 the plane is $bx + cy + az + d = 0$,which becomes $23x + 29y + 3z + d = 0$. Since the plane passes through the point $(1, 1, 1)$,we substitute these coordinates into the equation: $23(1) + 29(1) + 3(1) + d = 0$ $23 + 29 + 3 + d = 0$ $55 + d = 0$ $d = -55$. Thus,the equation of the plane is $23x + 29y + 3z - 55 = 0$,or $23x + 29y + 3z = 55$.

$A$ 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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