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Find a point P on the line \(\frac{x+5}{1} = \frac{y+3}{4}+\frac{z-6}{-9}\) such that its distance from point Q(2, 4, –1) is 7 units. Also, find the equation of line joining P and Q.

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Any point P on the line

\(\frac{x+5}{1} = \frac{y+3}{4} = \frac{z-6}{-9} = \lambda\)

is \(P(\lambda - 5, 4\lambda - 3, 6-9\lambda)\)

\(\because\) it is given that the distance between point P and Q is 7 units.

\(\Rightarrow |PQ| = 7\) 

\(= |PQ|^2 = 49\)

Now \(\vec{PQ} = (\lambda - 7)\hat{i} + (4\lambda - 7)\hat{j} +(7-9\lambda)\hat{k}\)

\(\Rightarrow (\lambda - 7)^2 + (4\lambda - 7)^2+(7-9\lambda)^2 = 49\)

\(\Rightarrow \lambda = 1\)

\(\therefore\) point P will be (–4, 1, –3)

The equation of line joining (–4, 1, –3) and (2, 4, –1) will have direction ratio \(<6,3,2>\)

\(\therefore\) Line joining P & Q is

\(\frac{x-2}{6} = \frac{y-4}{3} = \frac{z+1}{2}\)

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