A plane passes through the point (3, 5, 7). I | vedclass

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(D) The given equation of the plane is $x + 3y + 2z = 9$. Dividing by $9$,we get $\frac{x}{9} + \frac{y}{3} + \frac{z}{9/2} = 1$. Thus,the intercepts

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$6x + 2y + 3z = 49$

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(D) The given equation of the plane is $x + 3y + 2z = 9$. Dividing by $9$,we get $\frac{x}{9} + \frac{y}{3} + \frac{z}{9/2} = 1$. Thus,the intercepts made by the plane with the coordinate axes are $a = 9$,$b = 3$,and $c = 9/2$. The direction ratios of the normal to the required plane are $(9, 3, 9/2)$. Multiplying by $2$,we can take the direction ratios as $(18, 6, 9)$,or simply $(6, 2, 3)$. The plane passes through the point $(3, 5, 7)$. The equation of the plane is $a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$. Substituting the values: $6(x - 3) + 2(y - 5) + 3(z - 7) = 0$. $6x - 18 + 2y - 10 + 3z - 21 = 0$. $6x + 2y + 3z - 49 = 0$. Therefore,the equation of the plane is $6x + 2y + 3z = 49$.

$A$ plane passes through the point $(3, 5, 7)$. If the direction ratios of its normal are equal to the intercepts made by the plane $x + 3y + 2z = 9$ with the coordinate axes,then the equation of that plane is

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