Find all numbers k?

Given:

#vec(u) = < 2, -1, 0 >#

#vec(v) = < 1, 2, -1 >#

Then:

#k vec(u) - vec(v) = k < 2, -1, 0 > - < 1, 2, -1 >#

#color(white)(k vec(u) - vec(v)) = < 2k-1, -k-2, 1 >#

So we want to solve:

#6 = abs(abs(k vec(u) - vec(v)))^2#

#color(white)(6) = (color(blue)(2k-1))^2+(color(blue)(-k-2))^2+(color(blue)(1))^2#

#color(white)(6) = 4k^2-color(red)(cancel(color(black)(4k)))+1+k^2+color(red)(cancel(color(black)(4k)))+4+1#

#color(white)(6) = 5k^2+6#

Hence:

#k = 0#

Another way...

Actually we could have noticed a couple of things and reasoned this in a different way...

#abs(abs(vec(v))) = abs(abs(< 1, 2, -1 >))#

#color(white)(abs(abs(vec(v)))) = sqrt((color(blue)(1))^2+(color(blue)(2))^2+(color(blue)(-1))^2)#

#color(white)(abs(abs(vec(v)))) = sqrt(1+4+1)#

#color(white)(abs(abs(vec(v)))) = sqrt(6)#

#vec(u) * vec(v) = < 2, -1, 0 > * < 1, 2, -1 >#

#color(white)(vec(u) * vec(v)) = (color(blue)(2))(color(blue)(1))+(color(blue)(-1))(color(blue)(2))+(color(blue)(0))(color(blue)(-1)) = 2-2+0 = 0#

So #vec(v)# and hence #-vec(v)# is already of the required length #sqrt(6)# and #vec(u)# is orthogonal to it.

So adding any non-zero multiple of #vec(u)# to #-vec(v)# will result in a longer vector.

Hence the only solution is #k=0#