JEE Main 2022 · 24 Jun · Shift 2, Mathematics question 14

Data. $\vec a_2-\vec a_1=(1,2,2)$, $\vec b_1=(2,3,\lambda)$, $\vec b_2=(1,4,5)$, so $\vec b_1\times\vec b_2=(15-4\lambda,\ \lambda-10,\ 5)$.

Distance. $(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)=5-2\lambda$, so
$$\frac{(5-2\lambda)^2}{(15-4\lambda)^2+(\lambda-10)^2+25}=\frac13.$$

Solve. $3(4\lambda^2-20\lambda+25)=17\lambda^2-140\lambda+350$ gives $\lambda^2-16\lambda+55=0$.

Sum of roots 16.

(A)