JEE Mains · Maths · STD 12 - 11. three dimension geometry
If the equation of the plane passing through the point \((1,1,2)\) and perpendicular to the line \(x-3 y+2 z-1=04 x-y+z\) is \(Ax + By + Cz =1\), then \(140( C - B + A )\) is equal to \(.........\).
- A \(14\)
- B \(13\)
- C \(12\)
- D \(15\)
Answer & Solution
Correct Answer
(D) \(15\)
Step-by-step Solution
Detailed explanation
\(x-3 y+2 z-1=0\) \(4 x-y+z=0\) \(\overrightarrow{ n }_1 \times \overrightarrow{ n }_2=\left|\begin{array}{ccc}\hat{ i } & \hat{ j } & \hat{ k } \\ 1 & -3 & 2 \\ 4 & -1 & 1\end{array}\right|\) \(=-\hat{ i }+7 \hat{ j }+11 \hat{ k }\) \(\operatorname{Dr}^5\) of normal to the…
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