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

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(D) Given plane equation: $x+3y+2z=9$. Dividing by $9$,we get $\frac{x}{9} + \frac{y}{3} + \frac{z}{4.5} = 1$. The intercepts on the coordinate axes a

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

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(D) Given plane equation: $x+3y+2z=9$. Dividing by $9$,we get $\frac{x}{9} + \frac{y}{3} + \frac{z}{4.5} = 1$. The intercepts on the coordinate axes are $a=9$,$b=3$,and $c=\frac{9}{2}$. These intercepts are the direction ratios of the normal to the required plane,so $\vec{n} = 9\hat{i} + 3\hat{j} + \frac{9}{2}\hat{k}$. The plane passes through the point $(3, 5, 7)$,so the position vector is $\vec{a} = 3\hat{i} + 5\hat{j} + 7\hat{k}$. The equation of the plane is given by $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$. Substituting the values: $(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (9\hat{i} + 3\hat{j} + \frac{9}{2}\hat{k}) = (3\hat{i} + 5\hat{j} + 7\hat{k}) \cdot (9\hat{i} + 3\hat{j} + \frac{9}{2}\hat{k})$. $9x + 3y + \frac{9}{2}z = 27 + 15 + \frac{63}{2}$. $9x + 3y + \frac{9}{2}z = 42 + 31.5 = 73.5$. Multiplying by $\frac{2}{3}$ to simplify: $6x + 2y + 3z = 49$.

$A$ 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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