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JEE Advanced · Mathematics · 31. 3D Geometry

If the distance between the plane \(A x-2 y+z=d\) and the plane containing the lines \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4} \quad\) and \(\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}\) is \(\sqrt{6}\), then \(|d|\) is

  1. A 9
  2. B 7
  3. C 6
  4. D 5
Verified Solution

Answer & Solution

Correct Answer

(C) 6

Step-by-step Solution

Detailed explanation

Equation of plane containing the given lines is \(\left|\begin{array}{ccc}x-1 & y-2 & z-3 \\ 2 & 3 & 4 \\ 3 & 4 & 5\end{array}\right|=0\)
\[
\begin{array}{rr}
\Rightarrow & (x-1)(-1)-(y-2)(-2) \\
& +(z-3)(-1)=0 \\
\Rightarrow & -x+1+2 y-4-z+3=0 \\
\Rightarrow & -x+2 y-z=0 \quad \ldots(i)
\end{array}
\]
Given plane is
\[
x-2 y+z=d
\]

Eqs. (i) and (ii) are parallel.
Clearly, \(\quad A=1\)
Now, distance between plane
\[
\begin{aligned}
& =\left|\frac{d}{\sqrt{1+4+1}}\right|=\sqrt{6} \\
& \Rightarrow \quad|d|=6 \\
&
\end{aligned}
\]
From JEE Advanced
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