Let a =4 i - j +3 k, b =10 i +2 j - k and a vector c be such that 2( a × b )+3( b × c )= 0. If a · c =15, then c ·( i + j -3 k ) is equal to:

Let $\displaystyle \overrightarrow{\mathrm{a}}=4 \hat{i}-\hat{j}+3 \hat{k}, \overrightarrow{\mathrm{~b}}=10 \hat{i}+2 \hat{j}-\hat{k}$ and a vector $\displaystyle \overrightarrow{\mathrm{c}}$ be such that $\displaystyle 2(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})+3(\overrightarrow{\mathrm{~b}} \times \overrightarrow{\mathrm{c}})=\overrightarrow{0}$. If $\displaystyle \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=15$, then $\displaystyle \overrightarrow{\mathrm{c}} \cdot(\hat{i}+\hat{j}-3 \hat{k})$ is equal to: