JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let the plane passing through the point \((-1,0,-2)\) and perpendicular to each of the planes \(2 x+y-\) \(z=2\) and \(x-y-z=3\) be \(a x+b y+c z+8=0\). then the value of \(a+b+c\) is equal to:
- A \(8\)
- B \(4\)
- C \(3\)
- D \(5\)
Answer & Solution
Correct Answer
(B) \(4\)
Step-by-step Solution
Detailed explanation
Normal of req. plane \((2 \hat{i}+\hat{j}-\hat{k}) \times(\hat{i}-\hat{j}-\hat{k})\) \(=-2 \hat{i}+\hat{j}-3 \hat{k}\) Equation of plane \(-2(x+1)+1(y-0)-3(z+2)=0\) \(-2 x+y-3 z-8=0\) \(2 x-y+3 z+8=0\) \(a+b+c=4\)
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