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Q. The distance of the point $(-1,-5,-10)$ from the point of intersection of line $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}$ and plane $x-y+z=5$ is ' $\lambda$ ' then find value of $[\lambda / 3]$ (where $[\,\,.\,\,]$ represents greatest integer function)

Three Dimensional Geometry

Solution:

Equation of plane $x-y+z=5$ ... (1)
image
equation of line $P C \frac{x-2}{3}=\frac{y+1}{4}$
$=\frac{z-2}{12}=r$
$x=3 r+2, y=4 r-1, z=12 r+2$
coordinate of $P(3 r+2,4 r-1,12 r+2)$
Point $P$ lies in plane (1)
$(3 r+2)-(4 r-1)+12 r+2=5$
$11 r+5=5$
$r=0$
coordinate of $P (2,-1,2)$
$AP =\sqrt{9+16+144}=13$