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Question

# The equation of the plane containing the straight line x−12=y+2−3=z5 and perpendicular to the plane x−y+z+2=0 is

A
3x2y+4z2=0
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B
2x4y3z+6=0
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C
2x+3y+z+4=0
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D
x+2y+z+4=0
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Solution

## The correct option is C 2x+3y+z+4=0x−12=y+2−3=z−05 is perpendicular to the plane x−y+z+2=0 Required plane is parallel to the vector →v1=2^i−3^j+5^k and →v2=^i−^j+^k ∴ Normal vector to the required plane is, →n=∣∣ ∣ ∣∣^i^j^k2−351−11 ∣∣ ∣ ∣∣ =^i(−3+5)−^j(2−5)+^k(−2+3) =2^i+3^j+^k Also, the plane contains the point (1,−2,0) ∴ Equation of plane is 2(x−1)+3(y+2)+1(z−0)=0 ⇒2x+3y+z+4=0

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