If a plane passing through the points (2,3,0) | vedclass

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(B) Let the equation of the plane be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$. Since the plane passes through $(2,3,0), (0,-5,2)$ and $(-2,0,3)$,

Vedclass · Accepted Answer

$\left(\frac{7}{3}, 0,0\right)$

Vedclass · Answer

(B) Let the equation of the plane be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$. Since the plane passes through $(2,3,0), (0,-5,2)$ and $(-2,0,3)$,we have: $1) \frac{2}{a} + \frac{3}{b} = 1$ $2) -\frac{5}{b} + \frac{2}{c} = 1$ $3) -\frac{2}{a} + \frac{3}{c} = 1$ Adding equations $(1)$ and $(3)$,we get: $\frac{3}{b} + \frac{3}{c} = 2 \Rightarrow \frac{1}{b} + \frac{1}{c} = \frac{2}{3} \Rightarrow \frac{1}{b} = \frac{2}{3} - \frac{1}{c} = \frac{2c-3}{3c}$. Substitute $\frac{1}{b}$ into equation $(2)$: $-5\left(\frac{2c-3}{3c}\right) + \frac{2}{c} = 1 \Rightarrow \frac{-10c + 15 + 6}{3c} = 1 \Rightarrow -10c + 21 = 3c \Rightarrow 13c = 21 \Rightarrow c = \frac{21}{13}$. Now,from equation $(3)$: $-\frac{2}{a} + 3\left(\frac{13}{21}\right) = 1 \Rightarrow -\frac{2}{a} + \frac{13}{7} = 1 \Rightarrow \frac{2}{a} = \frac{13}{7} - 1 = \frac{6}{7} \Rightarrow a = \frac{14}{6} = \frac{7}{3}$. Thus,the intercept $A$ on the $X$-axis is $\left(\frac{7}{3}, 0, 0\right)$.

If a plane passing through the points $(2,3,0), (0,-5,2)$ and $(-2,0,3)$ meets the $X, Y, Z$-axes in $A, B, C$ respectively,then $A=$

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