JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let the mirror image of the point \((a, b\), c) with respect to the plane \(3 x-4 y+12 z+19=0\) be (a- \(6, \beta, \gamma)\). If \(a+b+c=5\), then \(7 \beta-9 \gamma\) is equal to
- A \(127\)
- B \(147\)
- C \(157\)
- D \(137\)
Answer & Solution
Correct Answer
(D) \(137\)
Step-by-step Solution
Detailed explanation
\(M =\left(a-3, \frac{\beta+b}{2}, \frac{\gamma+c}{2}\right)\) Since \(M\) lies on \(3 x+4 y+12 z+19=0\) \(\Rightarrow 6 a-4 b+12 c-4 \beta+12 \gamma+20=0 \text {. }\) Since \(PP'\) is parallel to normal of the plane then \(\frac{6}{3}=\frac{b-\beta}{-4}=\frac{c-\gamma}{12}\)…
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