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Question

The cartesian equation of the plane perpendicular to vector 3¯i2¯j2¯k and passing through the point 2¯i+3¯j¯k is

A
3x+2y+2z=2
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B
3x2y+2z=2
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C
3x+2y2z=2
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D
3x2y2z=2
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Solution

The correct option is D 3x2y2z=2
(ra).N=0a=2^i+3^j^kN=3^i2^j2^k(r2^i3^j+^k).(3^i2^j2^k)=0r=x^i+y^j+^k{(x2)i+(y3)^j+(z+1)^k}.(3i2^j2^k)=03(x2)2(y3)2(z+1)=03x62y+62z2=03x2y2z=2Ans.

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