MHT CET · Maths · Three Dimensional Geometry
The equation of the plane passing through the point \((1,1,1)\) and perpendicular to the planes \(2 x-y-2 z=5\) and \(3 x-6 y+2 z=7\) is
- A \(14 x+10 y+9 \mathrm{z}=13\)
- B \(14 x+10 y+9 z=33\)
- C \(14 x+10 y+9 z=-15\)
- D \(14 x+10 y+9 z=-33\)
Answer & Solution
Correct Answer
(D) \(14 x+10 y+9 z=-33\)
Step-by-step Solution
Detailed explanation
Required plane is perpendicular to planes \(2 x-y-2 z=5\) and \(3 x-6 y+2 z=7\)
Equation of required plane is
\(\begin{aligned}
& \left|\begin{array}{ccc}
x-1 & y-1 & z-1 \\
2 & -1 & -2 \\
3 & -6 & 2
\end{array}\right|=0 \\
& \Rightarrow(x-1)(-14) \div(y-1)(10) \\
& \quad+(z-1)(-12+3)=0 \\
& \Rightarrow-14 x+14-10 y+10-9 z+9=0 \\
& \Rightarrow 14 x+10 y+9 z=-33
\end{aligned}\)
Equation of required plane is
\(\begin{aligned}
& \left|\begin{array}{ccc}
x-1 & y-1 & z-1 \\
2 & -1 & -2 \\
3 & -6 & 2
\end{array}\right|=0 \\
& \Rightarrow(x-1)(-14) \div(y-1)(10) \\
& \quad+(z-1)(-12+3)=0 \\
& \Rightarrow-14 x+14-10 y+10-9 z+9=0 \\
& \Rightarrow 14 x+10 y+9 z=-33
\end{aligned}\)
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