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Obtain the point of intersection of the line joining the points {\bar  {i}}-2{\bar  {j}}-{\bar  {k}},2{\bar  {i}}+3{\bar  {j}}+{\bar  {k}}\, with the plane through the points 2{\bar  {i}}+{\bar  {j}}-3{\bar  {k}},4{\bar  {i}}-{\bar  {j}}+2{\bar  {k}}\, and 3{\bar  {i}}+{\bar  {k}}\,

The vector equation of the line joining the points {\bar  {i}}-2{\bar  {j}}-{\bar  {k}},2{\bar  {i}}+3{\bar  {j}}+{\bar  {k}}\, is {\bar  {r}}=(1-t)({\bar  {i}}-2{\bar  {j}}-{\bar  {k}})+t(2{\bar  {i}}+3{\bar  {j}}+{\bar  {k}})=(1+t){\bar  {i}}+(5t-2){\bar  {j}}+(2t-1){\bar  {k}}\,

Vector equation of the plane through the given points is {\bar  {r}}=(1-s-p)(2{\bar  {i}}+{\bar  {j}}-3{\bar  {k}})+s(4{\bar  {i}}-{\bar  {j}}+2{\bar  {k}})+p(3{\bar  {i}}+{\bar  {k}})=(2+2s+p){\bar  {i}}+(1-2s-p){\bar  {j}}+(5s+4p-3){\bar  {k}}\,

Solving the above two equations, 2+2s+p=1+t,2s+p+5t=3,5s+4p-2t=2\,

Solving the first two of the above three equations,we get t={\frac  {2}{3}}\,

Hence the point of intersection is {\bar  {r}}={\frac  {1}{3}}(5{\bar  {i}}+4{\bar  {j}}+{\bar  {k}})\,

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