If $$\vec{a}, \vec{b}, \vec{c}$$ are non-coplanar vectors and $$\lambda$$ is a real number then $$[\lambda(\vec{a} + \vec{b}) \, \lambda^2 \vec{b} \, \lambda \vec{c}] = [\vec{a} \, \vec{b} + \vec{c} \, \vec{b}]$$ for

The scalar triple product is defined as $$[\vec{p}\,\vec{q}\,\vec{r}] = \vec{p} \cdot (\vec{q}\times\vec{r})$$ and has the following two properties:
(i) It is linear in each of the three vectors.
(ii) If any two vectors are the same, the value is zero.

Calculate the left-hand side (LHS):

$$[\lambda(\vec{a}+\vec{b})\,\,\lambda^{2}\vec{b}\,\,\lambda\vec{c}] = \lambda\cdot\lambda^{2}\cdot\lambda\;[\vec{a}+\vec{b},\;\vec{b},\;\vec{c}] = \lambda^{4}\,[\vec{a}+\vec{b},\;\vec{b},\;\vec{c}]$$

Now split the triple product inside:

$$[\vec{a}+\vec{b},\;\vec{b},\;\vec{c}] = [\vec{a},\vec{b},\vec{c}] + [\vec{b},\vec{b},\vec{c}]$$

The second term is zero because two vectors are identical, hence

$$[\vec{a}+\vec{b},\;\vec{b},\;\vec{c}] = [\vec{a},\vec{b},\vec{c}]$$

Therefore

$$\text{LHS} = \lambda^{4}\,[\vec{a},\vec{b},\vec{c}]$$

Next, calculate the right-hand side (RHS):

$$[\vec{a},\;\vec{b}+\vec{c},\;\vec{b}] = [\vec{a},\vec{b},\vec{b}] + [\vec{a},\vec{c},\vec{b}]$$

Again, the first term is zero, so

$$\text{RHS} = [\vec{a},\vec{c},\vec{b}]$$

An interchange of the last two vectors changes the sign of a scalar triple product, hence

$$[\vec{a},\vec{c},\vec{b}] = -[\vec{a},\vec{b},\vec{c}]$$

Set LHS equal to RHS:

$$\lambda^{4}\,[\vec{a},\vec{b},\vec{c}] \;=\; -[\vec{a},\vec{b},\vec{c}]$$

The vectors $$\vec{a},\vec{b},\vec{c}$$ are non-coplanar, so $$[\vec{a},\vec{b},\vec{c}] \neq 0$$. Divide both sides by this non-zero quantity:

$$\lambda^{4} = -1$$

For real $$\lambda$$, the expression $$\lambda^{4}$$ is always non-negative, whereas the right-hand side is negative. Therefore the equation has no real solution.

Hence, there is no real value of $$\lambda$$ satisfying the given equality.

Option B which is: no value of $$\lambda$$