JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let the plane \(ax \,\,+\,\,by \,\,+c z=d\) pass through \((2,3,-5)\) and is perpendicular to the planes \(2 x + y -5 z =10\) and \(3 x+5 y-7 z=12\). If \(a, b, c, d\) are integers \(d>0\) and gcd \((lal, |b|,|c|, d)\) \(=1\), then the value of \(a+7 b+c+20 \,d\) is equal to
- A \(18\)
- B \(20\)
- C \(24\)
- D \(22\)
Answer & Solution
Correct Answer
(D) \(22\)
Step-by-step Solution
Detailed explanation
\(DR'S\) normal of plane \(\left|\begin{array}{ccc}\hat{ i } & \hat{ j } & \hat{ k } \\ 2 & 1 & -5 \\ 3 & 5 & -7\end{array}\right|=18 \hat{ i }-\hat{ j }+7 \hat{ k }\) \(\therefore eq ^{ a }\) of plane \(18 x - y +7 z = d\) It passes through \((2,3,-5)\)…
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