Dear Student,
Please find below the solution to the asked query:

We are looking for perpendicular distance.We have:P=-2i^+4j^-5k^L:x+33=y-45=z+86=tDirection vector is given by:DR=3i^+5j^+6k^Any general point on line will be:Q=3t-3i^+5t+4j^+6t-8k^PQ=3t-3i^+5t+4j^+6t-8k^--2i^+4j^-5k^=3t-3+2i^+5t+4-4j^+6t-8+5k^PQ=3t-1i^+5tj^+6t-3k^PQ will be perpendicular to DR, hencePQ.DR=03t-1i^+5tj^+6t-3k^.3i^+5j^+6k^=033t-1+25t+66t-3=09t-3+25t+36t-18=070t=21t=2170Hence point Q3t-3, 5t+4, 6t-8=6370-3,10570+4, 12670-8=-14770,38570, -43470Now required distance can be found by distance formula.PQ=-14770+22+38570-42+-43470+52=-7702+105702+-84702=49+11025+705670=1813070

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