JEE Mains · Maths · STD 12 - 11. three dimension geometry
If a plane passes through the points \((-1, k , 0),(2, k ,-1)\), \((1,1,2)\) and is parallel to the line \(\frac{x-1}{1}=\frac{2 y+1}{2}\) \(=\frac{ z +1}{-1}\), then the value of \(\frac{ k ^2+1}{( k -1)( k -2)}\) is
- A \(\frac{17}{5}\)
- B \(\frac{5}{17}\)
- C \(\frac{6}{13}\)
- D \(\frac{13}{6}\)
Answer & Solution
Correct Answer
(D) \(\frac{13}{6}\)
Step-by-step Solution
Detailed explanation
\(\frac{x-1}{1}=\frac{2 y+1}{2}=\frac{z+1}{-1}\) \(\frac{x-1}{1}=\frac{y+\frac{1}{2}}{1}=\frac{z+1}{-1}\) \(\text { Points : } A (-1, k , 0), B (2, k ,-1), C (1,1,2)\) \(\overrightarrow{ CA }=-2 \hat{ i }+( k -1) \hat{ j }-2 \hat{ k }\)…
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