The number of real values of $$\lambda$$ for which the system of linear equations, $$2x + 4y - \lambda z = 0$$, $$4x + \lambda y + 2z = 0$$ and $$\lambda x + 2y + 2z = 0$$, has infinitely many solutions, is:

For a homogenous system of equations to have infinitely many solutions, the determinant of the coefficients must be zero.

$$2\left(2\cdot\lambda\text{}-2\cdot2\right)-4\left(4\cdot2-\lambda\text{}\cdot2\right)+\left(-\lambda\text{}\right)\left(4\cdot2-\lambda\text{}\cdot\right)=0$$

$$4\lambda\text{}-8-32+8\lambda\text{}-8\lambda\text{}+\lambda\text{}^3=0$$

$$\therefore\ \lambda\text{}^3+4\lambda\text{}-40=0$$

$$f\left(\lambda\text{}\right)=\lambda\text{}^3+4\lambda\text{}-40$$

$$For\ \lambda\text{}=2,\ f\left(2\right)=8+8-40=-24<0$$

$$For\ \lambda\text{}=3,\ f\left(3\right)=27+12-40=-1<0$$

$$For\ \lambda\text{}=4,\ f\left(4\right)=64+16-40=40>0$$

$$f'\left(\lambda\text{}\right)=3\lambda\text{}^2+4$$

$$\therefore$$ The equation $$f\left(\right)$$ has only 1 real root between $$\lambda\text{}=3\ \&\ \lambda\text{}=4$$.

$$\therefore\ $$ The given system of equations has infinitely many solutions only for 1 real value of $$\lambda\text{}$$