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Question

Find the equation of the plane passing through the straight line x12=y+23=z5 and perpendicular to the plane xy+z+2=0.

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Solution

Line:x12=y+23=z5,plane;xy+z=0a=^i+(2)^j+0^k
is a point on the given line where direction is
b=2^i3^j+5^kc=^i^j+^k is normal to given plane.
The new plane contains line and is to plane.
Its normal to the direction line & also to normal of given plane, if n be the normal of required plane,
n=a×b=(^i+(2)^j+0^k)×(2^i3^j+5^k)=2^i3^j^k
d.r's of normal of new planev
Equation of plane through (1,2,0)&d.r's of normal (2,3,1) is,
2(x1)+3(y+2)+(z0)=02x+3y+z+4=0requiredequationofplane.

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