JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let \(L\) be a line obtained from the intersection of two planes \(x+2 y+z=6\) and \(y+2 z=4\) If point \(P (\alpha, \beta, \gamma)\) is the foot of perpendicular from \((3,2,1)\) on \(L ,\) then the value of \(21(\alpha+\beta+\gamma)\) equals ...... .
- A \(142\)
- B \(68\)
- C \(136\)
- D \(102\)
Answer & Solution
Correct Answer
(D) \(102\)
Step-by-step Solution
Detailed explanation
\(x+2 y+z=6\) \((y+2 z=4) \times 2\) ___________________ \(x-3 z=-2\) \(\Rightarrow x=3 z-2 \Rightarrow y=4-2 z\) \(\frac{x+2}{3}=z \quad \frac{y-4}{-2}=z\) \(\Rightarrow\) line of intersection of two planes is \(\frac{x+2}{3}=\frac{y-4}{-2}=z=\lambda\)…
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