Question

The distance between the point (3, 4, 5) and the point where the line $\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}$ meets the plane x + y + z = 17 is
(a) 1
(b) 2
(c) 3
(d) None of these

Answer 1

(c) 3

$$\begin{gathered} \text{The coordinates of any point on the given line are of the form} \\ \frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}=\lambda \\ \Rightarrow x=\lambda +3;y=2\lambda +4;z=2\lambda +5 \\ \text{So, the coordiantes of the point on the given line are}\left(\lambda +3,2\lambda +4,2\lambda +5\right)\text{. This point lies on the plane,} \\ x+y+z=17 \\ \Rightarrow \lambda +3+2\lambda +4+2\lambda +5=17 \\ \Rightarrow 5\lambda =5 \\ \Rightarrow \lambda =1 \\ \mathrm{So,\; the\; coordi}\text{na}\mathrm{tes\; of\; the\; point}\text{are} \\ \left(\lambda +3,2\lambda +4,2\lambda +5\right) \\ =\left(1+3,2\left(1\right)+4,2\left(1\right)+5\right) \\ =\left(4,6,7\right) \\ \text{Now, the distance between the points}\left(4,6,7\right)\text{and}\left(3,4,5\right)\text{is} \\ \sqrt{{\left(3-4\right)}^2+{\left(4-6\right)}^2+{\left(5-7\right)}^2} \\ =\sqrt{1+4+4} \\ =3\text{units} \end{gathered}$$
So, the answer is (c).