Question

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)

Answer 1

Answer: 4.00

Equation of plane $x-y+z=5$ ... (1)

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