according to ques,
The coordinates of any point Q on the line are:
\[x\text{ }=\text{ }-2\lambda \text{ }+\text{ }4,\text{ }y\text{ }=\text{ }6\lambda \text{ }and\text{ }z\text{ }=\text{ }-3\lambda \text{ }+\text{ }1\]
the given point is \[P\left( 2,\text{ }3,\text{ }-8 \right)\]
The direction ratios of PQ are:
\[-2\lambda \text{ }+\text{ }4\text{ }\text{ }2,\text{ }6\lambda \text{ }\text{ }3,\text{ }-3\lambda \text{ }+\text{ }1\text{ }+\text{ }8\text{ }i.e.\text{ }-2\lambda \text{ }+\text{ }2,\text{ }6\lambda \text{ }\text{ }3,\text{ }-3\lambda \text{ }+\text{ }9\]
And the direction ratios of the given line are:
\[-2,\text{ }6,\text{ }-3.\]
If PQ ⊥ line,
\[-2\left( -2\lambda \text{ }+\text{ }2 \right)\text{ }+\text{ }6\left( 6\lambda \text{ }\text{ }3 \right)\text{ }\text{ }3\left( -3\lambda \text{ }+\text{ }9 \right)\text{ }=\text{ }0\]
Or,
\[4\lambda \text{ }\text{ }4\text{ }+\text{ }36\lambda \text{ }\text{ }18\text{ }+\text{ }9\lambda \text{ }\text{ }27\text{ }=\text{ }0\]
Or,
\[49\lambda \text{ }\text{ }49\text{ }=\text{ }0\]
Or,
\[\lambda \text{ }=\text{ }1\]
Now, the foot of the perpendicular is:
\[-2\left( 1 \right)\text{ }+\text{ }4,\text{ }6\left( 1 \right),\text{ }-3\left( 1 \right)\text{ }+\text{ }1\]
\[i.e.\text{ }2,\text{ }6,\text{ }-2\]
therefore, the distance PQ is:
