Question
The distance of the point $$\left( {1,\, - 2,\,3} \right)$$ from the plane $$x - y + z = 5$$ measured parallel to the line $$\frac{x}{2} = \frac{y}{3} = \frac{{z - 1}}{{ - 6}}{\text{ is :}}$$
A.
$$1$$
B.
$$2$$
C.
$$4$$
D.
$$2\sqrt 3 $$
Answer :
$$1$$
Solution :
Equation of the line through $$\left( {1,\, - 2,\,3} \right)$$ parallel to the line $$\frac{x}{2} = \frac{y}{3} = \frac{{z - 1}}{{ - 6}}$$ is
$$\frac{{x - 1}}{2} = \frac{{y + 2}}{3} = \frac{{z - 3}}{{ - 6}} = r\,\,\,\,\left( {{\text{say}}} \right)......\left( 1 \right)$$
Then any point on $$\left( 1 \right)$$ is $$\left( {2r + 1,\,3r - 2,\, - 6r + 3} \right)$$
If this point lies on the plane $$x - y + z = 5$$ then
$$\left( {2r + 1} \right) - \left( {3r - 2} \right) + \left( { - 6r + 3} \right) = 5 \Rightarrow r = \frac{1}{7}$$
Hence the point is $$\left( {\frac{9}{7},\, - \frac{{11}}{7},\,\frac{{15}}{7}} \right)$$
Distance between $$\left( {1,\, - 2,\,3} \right)$$ and $$\left( {\frac{9}{7},\, - \frac{{11}}{7},\,\frac{{15}}{7}} \right)$$
$$\eqalign{
& = \sqrt {\left( {\frac{4}{{49}} + \frac{9}{{49}} + \frac{{36}}{{49}}} \right)} \cr
& = \sqrt {\left( {\frac{{49}}{{49}}} \right)} \cr
& = 1 \cr} $$