JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let \(P\) be a plane containing the line \(\frac{x-1}{3}=\frac{y+6}{4}=\frac{z+5}{2}\) and parallel to the line \(\frac{x-3}{4}=\frac{y-2}{-3}=\frac{z+5}{7} .\) If the point \((1,-1, \alpha)\) lies on the plane \(P\), then the value of \(|5 \alpha|\) is equal to ....... .
- A \(42\)
- B \(32\)
- C \(38\)
- D \(45\)
Answer & Solution
Correct Answer
(C) \(38\)
Step-by-step Solution
Detailed explanation
Equation of plane is \(\left|\begin{array}{ccc}x-1 & y+6 & z+5 \\ 3 & 4 & 2 \\ 4 & -3 & 7\end{array}\right|=0\) Now \((1,-1 ,\alpha\) ) lies on it so…
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