WBJEE · Maths · Three Dimensional Geometry
If the distance between the plane \(a x-2 y+z=k\) and the plane containing the lines \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\) and \(\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}\) is \(\sqrt{6}\), then \(|k|\) is
- A 36
- B 12
- C 6
- D \(2 \sqrt{3}\)
Answer & Solution
Correct Answer
(C) 6
Step-by-step Solution
Detailed explanation
Hint: Equation of plane \(\begin{aligned} & \left|\begin{array}{ccc} x-1 & y-2 & z-3 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{array}\right|=0 \\ & \Rightarrow x-2 y+z=0 \quad ---(1) \end{aligned}\) & given \(\alpha x-2 y+z=k\) ---(2) (1) // (2) \(\Rightarrow \alpha=1, k \neq 0\)…
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