1.Line l1l_1l1 through (2,6,2)(2,6,2)(2,6,2) perpendicular to 2x+y−2z=102x+y-2z=102x+y−2z=10. Shortest distance between l1l_1l1 and x+12=y+4−3=z2\frac{x+1}{2}=\frac{y+4}{-3}=\frac{z}{2}2x+1=−3y+4=2z isa.7b.19/319/319/3c.19/219/219/2d.9Login to continueOnly logged in users canattempt or see the solution.
