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Find the equation of the line passing through the point `(-1,2,3)` and perpendicular to the lines `x/2=(y-1)/(-3)=(z+2)/(-2)a n d(x+3)/(-1)=(y+3)/2=(z-1)/3dot`

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The line through point (-1, 2, 3) is perpendicular to the lines `(x)/(2)=(y-1)/(-3)=(2+2)/(-2) and (x+3)/(-1)= (y+3)/(-1)=(y+3)/(2)=(z-1)/(3)`.
Therefore, the line is along the vector `|{:(hati,,hatj,,hatk),(2,,-3,,-2),(-1,,2,,3):}|`
or `-5hati-4hatj+hatk`.
Hence, equation of the line is `(x+1)/(5)=(y-2)/(4)=(z-3)/(-1)`.
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