Question
The distance of the point $$\left( {1,\,3,\, - 7} \right)$$ from the plane passing through the point $$\left( {1,\, - 1,\, - 1} \right),$$ having normal perpendicular to both the lines $$\frac{{x - 1}}{1} = \frac{{y + 2}}{{ - 2}} = \frac{{z - 4}}{3}$$ and $$\frac{{x - 2}}{2} = \frac{{y + 1}}{{ - 1}} = \frac{{z + 7}}{{ - 1}},$$ is :
A.
$$\frac{{10}}{{\sqrt {74} }}$$
B.
$$\frac{{20}}{{\sqrt {74} }}$$
C.
$$\frac{{10}}{{\sqrt {83} }}$$
D.
$$\frac{5}{{\sqrt {83} }}$$
Answer :
$$\frac{{10}}{{\sqrt {83} }}$$
Solution :
Let the plane be $$a\left( {x - 1} \right) + b\left( {y + 1} \right) + c\left( {z + 1} \right) = 0$$
Normal vector
\[\left| \begin{array}{l}
\hat i\,\,\,\,\,\,\,\,\,\,\,\hat j\,\,\,\,\,\,\,\,\,\,\,\,\hat k\\
1\,\,\, - 2\,\,\,\,\,\,\,\,\,\,3\\
2\,\,\, - 1\,\, - 1
\end{array} \right| = 5\hat i + 7\hat j + 3\hat k\]
So plane is
$$\eqalign{
& 5\left( {x - 1} \right) + 7\left( {y + 1} \right) + 3\left( {z + 1} \right) = 0 \cr
& \Rightarrow 5x + 7y + 3z + 5 = 0 \cr} $$
Distance of point $$\left( {1,\,3,\, - 7} \right)$$ from the plane is
$$\frac{{5 + 21 - 21 + 5}}{{\sqrt {25 + 49 + 9} }} = \frac{{10}}{{\sqrt {83} }}$$