JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let the equation of the plane passing through the line \(x-2 y-z-5=0=x+y+3 z-5\) and parallel to the line \(x + y +2 z -7=0=2 x +3 y + z -2\) be \(a x+b y+c z=65\). Then the distance of the point \((a, b, c)\) from the plane \(2 x+2 y-z+16=0\) is \(..........\).
- A \(8\)
- B \(9\)
- C \(10\)
- D \(11\)
Answer & Solution
Correct Answer
(B) \(9\)
Step-by-step Solution
Detailed explanation
Equation of plane is \((x-2 y-z-5)+b(x+y+3 z-5)=0\) \(\left|\begin{array}{lll}1+b & -2+b & -1+3 b \\ 1 & 1 & 2 \\ 2 & 3 & 1\end{array}\right|=0\) \(\Rightarrow b =12\) \(\therefore \text { plane is } 13 x+10 y +35 z =65\) Distance from given point to plane \(=9\)
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