(i) The statement that $\overset{\rightarrow}{a}, \overset{\rightarrow}{b}, \overset{\rightarrow}{c}$ and $\overset{\rightarrow}{d}$ must each be a null vector, if $\overset{\rightarrow}{a}+\overset{\rightarrow}{b}+ \overset{\rightarrow}{c}+\overset{\rightarrow}{d}= 0$ ; is not correct. It is because $\overset{\rightarrow}{a}+\overset{\rightarrow}{b}+ \overset{\rightarrow}{c}+\overset{\rightarrow}{d}$ can be zero in many ways other than that $\overset{\rightarrow}{a}, \overset{\rightarrow}{b}, \overset{\rightarrow}{c}$and$\overset{\rightarrow}{d}$ must each be a null vector. (ii) Since $\overset{\rightarrow}{a}+\overset{\rightarrow}{b}+ \overset{\rightarrow}{c}+\overset{\rightarrow}{d}= 0$ ; $\left( \overset{\rightarrow}{a}+\overset{\rightarrow}{c}\right)$ = − $\left( \overset{\rightarrow}{b}+\overset{\rightarrow}{d}\right)$ Thus, vector $\left( \overset{\rightarrow}{a}+\overset{\rightarrow}{c}\right)$ is equal to negative of vector $\left( \overset{\rightarrow}{b}+\overset{\rightarrow}{d}\right)$ and hence the statement that magnitudeof $\left( \overset{\rightarrow}{a}+\overset{\rightarrow}{c}\right)$ is equal to the magnitude of $\left( \overset{\rightarrow}{b}+\overset{\rightarrow}{d}\right)$ is correct. (iii) $\text{ since } \overset{\rightarrow}{a}+ \overset{\rightarrow}{b}+ \overset{\rightarrow}{c}+\ \overset{\rightarrow}{d}=0 ;$$\overset{\rightarrow}{a}=-\left(\ \overset{\rightarrow}{b}+\ \overset{\rightarrow}{c}+\ \overset{\rightarrow}{d}\right)$ Therefore, magnitude of vector $\overset{\rightarrow}{a}$ is equal to magnitudeof vector $\left( \overset{\rightarrow}{b}+\ \overset{\rightarrow}{c}+\ \overset{\rightarrow}{d}\right)$. The sum of the magnitudes of vectors $\overset{\rightarrow}{b}, \overset{\rightarrow}{c}$ and $\overset{\rightarrow}{d}$ may be greater than or equal to that of vector $\overset{\rightarrow}{a}$. Hence, the statement that the magnitude of $\overset{\rightarrow}{a}$ can never be greater than the sum of the magnitudes of $\overset{\rightarrow}{b}, \overset{\rightarrow}{c}$ and $\overset{\rightarrow}{d}$ is correct.(iv) $\text{ since } \overset{\rightarrow}{a}+\ \overset{\rightarrow}{b}+\ \overset{\rightarrow}{c}+\ \overset{\rightarrow}{d}=0 ;$$\left(\ \overset{\rightarrow}{b}+\ \overset{\rightarrow}{c}\right)+\ \overset{\rightarrow}{a}+\ \overset{\rightarrow}{d}=0$The resultant sum of the three vectors $\ \overset{\rightarrow}{b}+\ \overset{\rightarrow}{c}, \ \overset{\rightarrow}{a}$ and $\overset{\rightarrow}{d}$ can be zero only if $\ \overset{\rightarrow}{b}+\ \overset{\rightarrow}{c}$ is in the plane of $\ \overset{\rightarrow}{a}$ and $\ \overset{\rightarrow}{d}.$ In case, the vectors $\overset{\rightarrow}{a}$ and $\ \overset{\rightarrow}{d}$ are collinear, $\ \overset{\rightarrow}{b}+\ \overset{\rightarrow}{c}$ must be in line of $\overset{\rightarrow}{a}$ and $\ \overset{\rightarrow}{d}$Hence, the given statement is correct.