The lines `bar r=bar i+bar j-bar k+s(3bar i-bar j) and bar r=4bar i-bar k+t(2bar i +3bar k)`

Find the shortest distance between the skew lines . bar(r)=(6bar(i)+2bar(j)+2bar(k))+t(bar(i)-2bar(j)+2bar(k))andbar(r)=(-4bar(i)-bar(k))+s(3bar(i)-2bar(j)-2bar(k))

The shortest distance between the lines bar(r)=3bar(i)+5bar(j)+7bar(k)+lambda(bar(i)+2bar(j)+bar(k)) and bar(r)=-bar(i)-bar(j)+bar(k)+mu(7bar(i)-6bar(j)+bar(k)) is

bar (a) = 2bar (i) + 3bar (j) -bar (k), bar (b) = bar (i) + 2bar (j) -4bar (k), bar (c) = bar (i) + bar (j) + bar (k), bar (d) = bar (i) -bar (j) -bar (k) then

Find the vector area and area of the triangle with vertices bar(i) + bar(j)-bar(k), 2bar(i)- 3bar(j) + bar(k), 3 bar(i) + bar(j)- 2bar(k) .

The shortest distance the two lines bar(r)=2bar(i)-bar(j)-bar(k)+lambda(2bar(i)+bar(j)+2bar(k)) and bar(r)=(bar(i)+2bar(j)+bar(k))+mu(bar(i)-bar(j)+bar(k)) is

Find the distance between the planes bar(r) * ( 2 bar(i) - bar(j) + 3 bar(k)) = 4 and bar(r) * ( 6 bar(i) - 3 bar(j) + 9 bar(k)) + 13 = 0

The vectors 2bar(i)-3bar(j)+bar(k), bar(i)-2bar(j)+3bar(k), 3bar(i)+bar(j)-2bar(k)

Prove that vectors bar(a) = 2bar(i) -bar(j) + bar(k), bar(b) = bar(i) -bar(3j)-5bar(k) and bar(c ) = 3bar(i) - 4bar(j) -4bar(k) are coplanar.