$\text{ If }a=2\widehat{i}-\widehat{j}+\widehat{k},b=\widehat{i}+2\widehat{j}-3\widehat{k}\text{ and }c=3\widehat{i}-y\widehat{j}+5\widehat{k}\text{ are coplanar. The value of y is \_\_\_\_\_}$

$\text{ If the vectors are coplanar. Then, }a\cdot (b\times c)=0\text{. }$

$\begin{array}{l}\Rightarrow (2\widehat{i}-\widehat{j}+\widehat{k})\cdot \left|\begin{array}{ccc}\widehat{i}& \widehat{j}& \widehat{k}\\ 1& 2& -3\\ 3& -y& 5\end{array}\right|=0\\ \Rightarrow (2\tilde{i}-\tilde{j}+\tilde{k})\cdot \left\{\tilde{i}\right(10-3y)-\tilde{j}(5+9)+\tilde{k}(-y-6\left)\right\}=0\\ \Rightarrow (2\tilde{i}-\tilde{j}+\tilde{k})\cdot \left\{\right(10-3y)\tilde{i}-14\tilde{j}-(6+y\left)\tilde{k}\right\}=0\\ \Rightarrow 2(10-3y)+14-(6+y)=0\\ \Rightarrow 20-6y+14-6-y=0\\ \Rightarrow 28-7y=0\therefore y=4\end{array}$