JEE Mains · Maths · STD 12 - 11. three dimension geometry
Equation of the plane which passes through the point of intersection of lines \(\frac{{x - 1}}{3} = \frac{{y - 2}}{1} = \frac{{z - 3}}{2}\) and \(\frac{{x - 3}}{1} = \frac{{y - 1}}{2} = \frac{{z - 2}}{3}\) and has the largest distance from the origin is
- A \(7x +2y+4z= 54\)
- B \(3x+4y+ 5z= 49\)
- C \(4x+3y+5z = 50\)
- D \(5x+4y+3z = 57\)
Answer & Solution
Correct Answer
(C) \(4x+3y+5z = 50\)
Step-by-step Solution
Detailed explanation
Given equation of lines are \(\frac{x-1}{3}=\frac{y-2}{1}=\frac{z-3}{2}\) .......\((1)\) and \(\frac{x-3}{1}=\frac{y-1}{2}=\frac{z-2}{3}\) ....\((2)\) Any point on line \(( 1)\) is \(P\) \((3 \lambda+1, \lambda+2,2 \lambda+3)\) and on line \(( 2)\) is \(Q\)…
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