MHT CET · Maths · Three Dimensional Geometry
The equation of the plane passing through the point of intersection of the planes \(2 x-y+z-3=0\) and \(4 x-3 y+5 z+9=0\) and parallel to the lines \(\frac{x+1}{2}=\frac{y+3}{4}=\frac{z-3}{5}\) is \(\alpha x+\beta y+\gamma z+d=0\) Then \(\alpha+\beta+\gamma+\mathrm{d}=\)
- A \(48\)
- B \(-48\)
- C \(84\)
- D \(45\)
Answer & Solution
Correct Answer
(B) \(-48\)
Step-by-step Solution
Detailed explanation
Equation of plane: \((2x - y + z - 3) + \lambda(4x - 3y + 5z + 9) = 0\) Normal vector \(N = \langle 2 + 4\lambda, -1 - 3\lambda, 1 + 5\lambda \rangle\)
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