Find the equation of the plane through the line of intersection of r(2i-3j+4k)
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the line of intersection vector of r(2i-3j+4k)
=r (2i-3i+4k)=1
=R. (I-J) +4=0
=r.(2i-j+k)+8=0
=r.(2i - 3j + 4k) = 1
=r. ( i - j) + 4 = 0
= r.(2i - j + k) + 8 = 0.
=x - 1 = 2y - 4 = 3z - 21
= r-2i-3j-4k
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