Question
The distance of the point (1, 0, 2) from the point of intersection of the line $$\frac{{x - 2}}{3} = \frac{{y + 1}}{4} = \frac{{z - 2}}{{12}}$$ and the plane $$x-y+z=16,$$ is :
A.
$$3\sqrt {21} $$
B.
$$13$$
C.
$$2\sqrt {14} $$
D.
$$8$$
Answer :
$$13$$
Solution :
General point on given line $$ \equiv P\left( {3r + 2,\,4r - 1,\,12r + 2} \right)$$
Point $$P$$ must satisfy equation of plane
$$\eqalign{
& \left( {3r + 2} \right) - \left( {4r - 1} \right) + \left( {12r + 2} \right) = 16 \cr
& \Rightarrow 11r + 5 = 16 \cr
& \Rightarrow r = 1 \cr
& P\left( {3 \times 1 + 2,\,4 \times 1 - 1,\,12 \times 1 + 2} \right)\,\,\,\,\, = P\left( {5,\,3,\,14} \right) \cr} $$
distance between $$P$$ and (1, 0, 2)
$$D = \sqrt {{{\left( {5 - 1} \right)}^2} + {3^2} + {{\left( {14 - 2} \right)}^2}} = 13$$