The distance of the point $(7,5,2)$ from the plane $3 x+4 y+z-8=0$ measured parallel to the line…
The distance of the point $(7,5,2)$ from the plane $3 x+4 y+z-8=0$ measured parallel to the line $\frac{x-1}{3}=\frac{y-2}{6}=\frac{z+1}{2}$
$\sqrt{74}$ units
$\sqrt{47}$ units
6 units
7 units
Solution
Let $\quad P=(7,5,2)$
Eq. of line passing through $P$ and parallel to given line is
$\frac{x-7}{3}=\frac{y-5}{6}=\frac{z-2}{2}=r \text { (say) }$
Hence coordinates of any point on this line are $(3 r+7,6 r+5,2 r+2) \equiv Q($ say $)$
We have $3 x+4 y+z-8=0$
$\therefore 3(3 r+7)+4(6 r+5)+(2 r+2)-8=0$
$\therefore 9 r+24 r+2 r+21+20+2-8=0 \Rightarrow 35 r=-35 \Rightarrow r=-1$
$\therefore Q \equiv(-3+7,-6+5,-2+2)$ i.e. $(4,-1,0)$
Distance between $\mathrm{PQ} \equiv \sqrt{(7-4)^{2}+(5+1)^{2}+(2-0)^{2}}=\sqrt{9+36+4}=7$