JEE Mains · Maths · STD 12 - 11. three dimension geometry
If the foot of the perpendicular drawn from \((1,9\), 7) to the line passing through the point \((3,2,1)\) and parallel to the planes \(x+2 y+z=0\) and \(3 y-z=3\) is \((\alpha, \beta, \gamma)\), then \(\alpha+\beta+\gamma\) is equal to
- A \(-1\)
- B \(3\)
- C \(1\)
- D \(5\)
Answer & Solution
Correct Answer
(D) \(5\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{l}\text { Direction ratio of line }=\left|\begin{array}{ccc}\hat{ i } & \hat{ j } & \hat{ k } \\ 1 & 2 & 1 \\ 0 & 3 & -1\end{array}\right| \\ =\hat{ i }(-5)-\hat{ j }(-1)+\hat{ k }(3) \\ =-5 \hat{ i }+\hat{ j }+3 \hat{ k }\end{array}\)…
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