The equation of line passing through (3,-1,2) and perpendicular to the lines
`vecr=(hati+hatj-hatk)+lamda(2hati-2hatj+hatk)`
and `vecr=(2hati+hatj-3hatk) +mu(hati-2hatj+2hatk)` is

Explanation

The line ` bar r = ( hati + hatj - hatk ) + lamda ( 2 hati - 2 hatj + hatk ) ` is parallel to the vector ` bar b = 2 hati - 2 hatj + hatk ` and the line `bar r = (2hati + hatj - 3hat k ) + mu ( hat i - 2hatj + 2hatk )` is parallel to the vector `barc = hati - 2hatj + hat 2k` .
The vector perpendicular to the vectors `bar b` and `bar c ` is given by
`bar b xx bar c = |{:(hati ,, hatj,,hatk ) , ( 2,, -2,,1), ( 1,,-2,,2):}|`
`= hati (-4 + 2) - hatj( 4 -1 ) + hatk ( - 4 + 2)`
` = - 2hati - 3hatj - 2hatk`
since the required line is perpendicular to the given lines , it is perpendicular to both `bar b "and" bar c `.
`therefore` it is parallel to `bar b xx bar c `
The equation of the line passing through A `bar(a)` and parallel to `bar(b)xxbar(c)` is
`bar(r) = bar(a) + lamda(bar b xx lamda c)` where `lamda` is a scalar .
Here , `bar(a) = 3hati - hatj + 2hatk`
`therefore` the equation of the required line is
`bar(r) = ( 3 hat i - hatj + 2hatk ) `
or `bar(r) = ( 3hati - hatj + 2hatk ) + mu ( -2hati = 3hatj - 2hatk ) "where" mu = - lamda`.