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Q.
Equation of plane containing the line $\frac{x-1}{-3} = \frac {y-3}{2} = \frac{z-2}{1}$ and the point $(0 , - 3, 4 )$ is
Three Dimensional Geometry
Solution:
Equation of plane containing the line
$a(x-1) +b (y-3)+c (z-2)=0$ and $-3a+2b+c=0$
Using $(0, -3, 4)$, we have
$a(-1)+b (-6)+c (2)=0$
or $a+6b-2c=0$
and $3a-2b-c=0$
$\therefore \frac{a}{10} = \frac{-b}{-5}=\frac{c}{20}$
or $\frac{a}{2} = \frac{b}{1} =\frac{c}{4}$
$\therefore $ Required equation of plane is
$2(x-1)+y-3+4(z-2)=0$
$\Rightarrow 2x+y+4z=13$