JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let \(P\) be the plane passing through the line \(\frac{x-1}{1}=\frac{y-2}{-3}=\frac{z+5}{7}\) and the point \((2,4,-3)\). If the image of the point \((-1,3,4)\) in the plane \(P\) is \((\alpha, \beta, \gamma)\), then \(\alpha+\beta+\gamma\) is equal to
- A \(12\)
- B \(11\)
- C \(9\)
- D \(10\)
Answer & Solution
Correct Answer
(D) \(10\)
Step-by-step Solution
Detailed explanation
Equation of plane is given by \(\left|\begin{array}{ccc} x -1 & y -2 & z +5 \\ 1 & 2 & 2 \\ 1 & -3 & 7\end{array}\right|=0\) \(4 x-y-z=7\) \(\frac{\alpha+1}{4}=\frac{\beta-3}{-1}=\frac{\gamma-4}{-1}=\frac{-2(-4-3-4-7)}{16+1+1}=2\) \(\alpha=7, \beta=1, \gamma=2\)…
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