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A straight line passes through $(1,-2,3)$ and perpendicular to the plane $2 x+3 y-z=7$
Where does the line meet the plane?
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Where does the line meet the plane?
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The correct answer is:
$(3,1,2)$
Equation of line, $\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{-1}$ Let $\mathrm{P}(2 \mathrm{r}+1,3 \mathrm{r}-2,-\mathrm{r}+3)$ of the line meets the plane. Then, $2(2 \mathrm{r}+1)+3(3 \mathrm{r}-2)-(-\mathrm{r}+3)=0$
$4 r+2+9 r-6+r-3=7$
$14 r=14$
$\mathbf{r}=1$
$\mathrm{P}(3,1,2)$ meets the plane.
$4 r+2+9 r-6+r-3=7$
$14 r=14$
$\mathbf{r}=1$
$\mathrm{P}(3,1,2)$ meets the plane.
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