Consider the three planes P_{1}: 3x + 15y + 2 | vedclass

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(B) The equation of plane $P_{1}$ is $3x + 15y + 21z = 9$. Dividing by $3$,we get $x + 5y + 7z = 3$.The equation of plane $P_{2}$ is $x - 3y - z = 5$.

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$P_{1}$ and $P_{3}$ are parallel

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(B) The equation of plane $P_{1}$ is $3x + 15y + 21z = 9$. Dividing by $3$,we get $x + 5y + 7z = 3$.The equation of plane $P_{2}$ is $x - 3y - z = 5$.The equation of plane $P_{3}$ is $2x + 10y + 14z = 5$. Dividing by $2$,we get $x + 5y + 7z = \frac{5}{2}$.Two planes $a_{1}x + b_{1}y + c_{1}z = d_{1}$ and $a_{2}x + b_{2}y + c_{2}z = d_{2}$ are parallel if their normal vectors are proportional,i.e.,$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$.Comparing $P_{1}$ and $P_{3}$,the normal vectors are $(1, 5, 7)$ and $(1, 5, 7)$,which are identical. Since the constant terms $3$ and $\frac{5}{2}$ are different,the planes are parallel and distinct.Therefore,$P_{1}$ and $P_{3}$ are parallel.

Consider the three planes $P_{1}: 3x + 15y + 21z = 9$; $P_{2}: x - 3y - z = 5$; and $P_{3}: 2x + 10y + 14z = 5$. Then,which one of the following is true?

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