Six faces of a die are marked with the numbers 1, -1, 0, -2, 2 and 3.…

Total number of elementary events = $6^3$

Favourable number of elementary events

= Coeff. of $x^6$ in $(x + x^{-1} +x^0 + x^{-2} +x^2 + x^3)^3$

= Coeff. of $x^6$ in $\left(\frac{1+x+x^2+x^3+x^4+x^5}{x^2}\right)^3$

= Coeff. of $x^{12}$ in $(1+x+x^2 +x^3 +x^4 +x^5)^3$

= Coeff. of $x^{12}$ in $\left(\frac{1-x^6}{1-x}\right)^3$

= Coeff. of $x^{12}$ in $(1-x^6)^3 (1-x)^{-3}$

= Coeff. of $x^{12}$ in $(1-{^3C}_1x^6 +{^3C}_2x^{12} ....) (1-x)^{-3}$

= Coeff. of $x^{12}$ in $(1-x)^{-3} - {^3C}_1×$ Coeff. of $x^6$ in $(1-x)^{-3} +{^3C}_2×$ Coeff. of $x^0$ in $(1-x)^{-3}$

$= {^{12+3-1}C}_{3-1} - {^3C}_1×{^{6+3-1}C}_{3-1}+{^3C}_2$

$= {^{14}C}_2-{^3C}_1 ×{^8C}_2+{^3C}_2 = 91 - 84 + 3 = 10.$

Hence, required probability $=\frac{10}{6^3}=\frac{5}{108}.$