Answer:
Given equations of lines are \[\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\] and \[\frac{x+1}{2}=\frac{y+2}{3}=\frac{z+3}{-\,2}.\] Then, DR's of given two lines are respectively \[{{a}_{1}}=2,\] \[{{b}_{1}}=3,\] \[{{c}_{1}}=4\] and \[{{a}_{2}}=1,\] \[{{b}_{2}}=2,\] \[{{c}_{2}}=-\,2\] Now, consider \[{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}\] \[=2(1)+3(2)+4(-\,2)=2+6-8=8-8=0\] \[\because \] \[{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}=0\] Hence, the given lines are perpendicular. Hence proved.
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