Three unbiased dice are rolled and the numbers on them are noted as a, b, c. The probability that the planes $\mathrm{ax}+\mathrm{by}+\mathrm{cz}=0,\mathrm{bx}+\mathrm{cy}+\mathrm{az}=0,\mathrm{cx}+\mathrm{ay}+\mathrm{bz}=0$

The given 3 planes are passing through (0, 0, 0). The 3 planes intersect along a line if the given system of equations has a non-trivial solution. 
$\begin{array}{l}\iff \mathrm{\Delta }=\left|\begin{array}{lll}\mathrm{a}& \mathrm{b}& \mathrm{c}\\ \mathrm{b}& \mathrm{c}& \mathrm{a}\\ \mathrm{c}& \mathrm{a}& \mathrm{b}\end{array}\right|=0\\ \iff \mathrm{a}=\mathrm{b}=\mathrm{c}\text{ or }\mathrm{a}+\mathrm{b}+\mathrm{c}=0\end{array}$
But if a = b = c the given equations represent same plane and a+ b + c $\ne$ 0($\because$a, b, c are obtained by rolling a fair die for 3 times) $\because$P(3 planes intersecting along a line) = $\mathrm{P}\left(\mathrm{\varphi }\right)=0$ (3 planes intersect at only one point (0,0,0)) = P(A)
$=1-\mathrm{P}\left[\overline{\mathrm{A}}\right]=1-\mathrm{P}(\mathrm{a}=\mathrm{b}=\mathrm{c})=1-\frac{6}{216}=\frac{105}{108}$
If 3 planes have common point other than origin then they are coincidental.
$\therefore$P(3 places form a triangular prism) = 0