The position vectors of the three are <1,2,3>, <4,λ,6>, <4,6,9>, represented by vectors A, B, C.
For linear dependence: c1A+c2B+c3C=0 for some scalars c1, c2, c3. The zero vector 0 is <0,0,0>.
This gives us the equations represented:
R1: c1+4c2+4c3=0
R2: 2c1+λc2+6c3=0
R3: 3c1+6c2+9c3=0⇒c1+2c2+3c3=0
We need a non-trivial solution to these equations (at least one of the c's must be non-zero) for linear dependence.
Now let R2 become R2-2R1: (λ-8)c2-2c3=0; and R3 become R3-3R1: -6c2-3c3=0, making c3=-2c2.
Also (λ-8)c2-2c3=0=(λ-8)c2+4c2⇒λ-8+4=0 (if c2≠0), and λ=4.
COROLLARY
Let c2=1 (arbitrarily), then:
R1: c1+4+4c3=0
R2: 2c1+4+6c3=0⇒c1+2+3c3=0
R3: c1+2+3c3=0=R2 (creates a redundancy in the set of vectors)
R1-R2: 2+c3=0, c3=-2⇒c1=4⇒4A+B-2C=0:
4<1,2,3>+<4,4,6>-2<4,6,9>=<0,0,0>✔️
We've already seen that c3=-2c2, so if k=c2, c1=4k and 4k<1,2,3>+k<4,4,6>-2k<4,6,9>=<0,0,0> for all k≠0.