Correct option is (2) 1120
Using family of planes
\(2 \mathrm{x}+3 \mathrm{y}-\mathrm{z}-5=\mathrm{k}_{1}(\mathrm{x}+\alpha \mathrm{y}+3 \mathrm{z}+4)+\mathrm{k}_{2}(3 \mathrm{x}-\mathrm{y}+\beta \mathrm{z}-7)\)
\(2=\mathrm{k}_{1}+3 \mathrm{k}_{2}, 3=\mathrm{k}_{1} \alpha-\mathrm{k}_{2},-1=3 \mathrm{k}_{1}+\beta \mathrm{k}_{2},-5=4 \mathrm{k}_{1}-7 \mathrm{k}_{2}\)
On solving we get
\(k_{2}=\frac{13}{19}, k_{1}=\frac{-1}{19}, \alpha=-70, \beta=\frac{-16}{13}\)
\(13 \alpha \beta=13(-70)\left(\frac{-16}{13}\right)\)
\(=1120\)
