JEE Mains · Maths · STD 12 - 11. three dimension geometry
If the equation of plane passing through the mirror image of a point \((2,3,1)\) with respect to line \(\frac{x+1}{2}=\frac{y-3}{1}=\frac{z+2}{-1}\) and containing the line \(\frac{x-2}{3}=\frac{1-y}{2}=\frac{z+1}{1}\) is \(\alpha x+\beta y+\gamma z=24\), then \(\alpha+\beta+\gamma\) is equal to ..... .
- A \(20\)
- B \(19\)
- C \(18\)
- D \(21\)
Answer & Solution
Correct Answer
(B) \(19\)
Step-by-step Solution
Detailed explanation
Line \(\frac{x+1}{2}=\frac{y-3}{1}=\frac{z+2}{-1}\) \(\overline{ PM }=(2 \lambda-3, \lambda,-\lambda-3)\) \(\overline{ PM } \perp(2 \hat{ i }+\hat{ j }-\hat{ k })\) \(4 \lambda-6+\lambda+\lambda+3=0 \Rightarrow \lambda=\frac{1}{2}\)…
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