(a)OR(b)

Find the equation of the line passing through the point of intersection of the lines $\displaystyle \frac{\mathrm{x}}{1}=\frac{\mathrm{y}-1}{2}=\frac{\mathrm{z}-2}{3}$ and $\displaystyle \frac{\mathrm{x}-1}{0}=\frac{\mathrm{y}}{-3}=\frac{\mathrm{z}-7}{2}$ and perpendicular to these given lines.

$$l_{1}: \frac{\mathrm{x}}{1}=\frac{\mathrm{y}-1}{2}=\frac{\mathrm{z}-2}{3}=\lambda ; l_{2}: \frac{\mathrm{x}-1}{0}=\frac{\mathrm{y}}{-3}=\frac{\mathrm{z}-7}{2}=\muany point on $\displaystyle l_{1}$ is $\displaystyle (\lambda, 2 \lambda+1,3 \lambda+2)$ & any point on $\displaystyle l_{2}$ is $\displaystyle (1,-3 \mu, 2 \mu+7)$ If $\displaystyle l_{1}$ and $\displaystyle l_{2}$ intersect,\lambda=1,2 \lambda+1=-3 \mu \text { and } 3 \lambda+2=2 \mu+7 \Rightarrow \lambda=1 \text { and } \mu=-1Point of intersection of $\displaystyle l_{1}$ and $\displaystyle l_{2}$ is $\displaystyle (1,3,5)$. Let d.r.'s of required line be $\displaystyle \langle a, b, c\rangle$. Then,a+2 b+3 c=0 \text { and }-3 b+2 c=0 \Rightarrow \frac{a}{13}=\frac{b}{-2}=\frac{c}{-3}Required equation of line is $\displaystyle \frac{\mathrm{x}-1}{13}=\frac{\mathrm{y}-3}{-2}=\frac{\mathrm{z}-5}{-3}$ d.r's of CD are $\displaystyle \langle 1,-2,2\rangle$ ∴ d.r's of AB are $\displaystyle <1,-2,2>$