If $\mathbf{a, b}$ and $\mathbf{c}$ are three vectors such that $\mathbf{a + b + c = 0}$ and $|\mathbf{a}| = 2, |\mathbf{b}| = 3$ and $|\mathbf{c}| = 5$, then the value of $\mathbf{a \cdot b + b \cdot c + c \cdot a}$ is

$-19$

The correct answer is Option (3) → $-19$ ##

Here, $\mathbf{a + b + c = 0}$ and $a^2 = 4, b^2 = 9, c^2 = 25$

or $\mathbf{a \cdot a} = 4, \mathbf{b \cdot b} = 9$ and $\mathbf{c \cdot c} = 25$

$∴(\mathbf{a + b + c}) \cdot (\mathbf{a + b + c}) = 0$

$\Rightarrow a^2 + \mathbf{a \cdot b + a \cdot c + b \cdot a} + b^2 + \mathbf{b \cdot c + c \cdot a + c \cdot b} + c^2 = 0$

$\Rightarrow a^2 + b^2 + c^2 + 2(\mathbf{a \cdot b + b \cdot c + c \cdot a}) = 0 \quad [∵\mathbf{a \cdot b = b \cdot a}]$

$\Rightarrow 4 + 9 + 25 + 2(\mathbf{a \cdot b + b \cdot c + c \cdot a}) = 0$

$\Rightarrow \mathbf{a \cdot b + b \cdot c + c \cdot a} = \frac{-38}{2} = -19$