Question:
The distance of the point $\mathrm{P}(3,4,4)$ from the point of intersection of the line joining the points. $\mathrm{Q}(3,-4,-5)$ and $\mathrm{R}(2,-3,1)$ and the plane $2 x+y+z=7$, is equal to
Solution:
$\overrightarrow{\mathrm{QR}}:-\frac{\mathrm{x}-3}{1}=\frac{\mathrm{y}+4}{-1}=\frac{\mathrm{z}+5}{-6}=\mathrm{r}$
$\Rightarrow(x, y, z) \equiv(r+3,-r-4,-6 r-5)$
Now, satisfying it in the given plane.
We get $r=-2$.
so, required point of intersection is $\mathrm{T}(1,-2,7)$.
Hence, $\mathrm{PT}=7 .$
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