Q41 / 66 · JEE Advanced 2008 (Paper 2) · Mathematics
Paragraph:
Consider the lines: $L_1: \frac{x+1}{3}=\frac{y+2}{1}=\frac{z+1}{2}, L_2: \frac{x-2}{1}=\frac{y+2}{2}=\frac{z-3}{3}$Question:
The distance of the point $(1,1,1)$ from the plane passing through the point $(-1,-2,-1)$ and whose normal is perpendicular to both the lines $L_1$ and $L_2$, is
- A$2 / \sqrt{75}$
- B$7 / \sqrt{75}$
- Correct answerC$13 / \sqrt{75}$
- D$23 / \sqrt{75}$
Solution
The equation of the plane passing through the point $(-1,-2,-1)$ and whose normal is perpendicular to both the given lines $L_1$ and $L_2$ may be written as
$$
\begin{array}{rlrl}
\Rightarrow \quad(x+1)+7(y+2)-5(z+1) & =0 \\
& x+7 y-5 z+10 & =0
\end{array}
$$
The distance of the point $(1,1,1)$ from the plane
$$
=\left|\frac{1+7-5+10}{\sqrt{1+49+25}}\right|=\frac{13}{\sqrt{75}}
$$
Hence, the correct answer is option: 'C'.