ExamBro
ExamBro
JEE Mains · Maths · STD 12 - 11. three dimension geometry

Let \((\alpha, \beta, \gamma)\) be the foot of perpendicular from the point \((1,2,3)\) on the line \(\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}\). then \(19(\alpha+\beta+\gamma)\) is equal to :

  1. A \(102\)
  2. B \(101\)
  3. C \(99\)
  4. D \(100\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(101\)

Step-by-step Solution

Detailed explanation

Let foot \(P(5 k-3,2 k+1,3 k-4)\) DR's \(\rightarrow\) AP: \(5 \mathrm{k}-4,2 \mathrm{k}-1,3 \mathrm{k}-7\) DR's \(\rightarrow\) Line: \(5,2,3\) Condition of perpendicular lines \((25 k-20)+(4 k-2)+(9 k-21)=0\) Then \(\mathrm{k}=\frac{43}{38}\) Then…
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app