A line passes through A(4, - 6, - 2) and B(16, - 2, 4) . The point...

Given,

A line passes through $A\left(4, - 6, - 2\right)$ and $B\left(16, - 2, 4\right)$,

So, the equation of line will be $\dfrac{x - 4}{12} = \dfrac{x + 6}{4} = \dfrac{z + 2}{6}$

$⇒ \dfrac{x - 4}{6} = \dfrac{x + 6}{2} = \dfrac{z + 2}{3}$

So, the direction cosines will be $\dfrac{6}{\sqrt{{6}^{2} + {2}^{2} + {3}^{2}}}, \dfrac{2}{\sqrt{{6}^{2} + {2}^{2} + {3}^{2}}}, \dfrac{3}{\sqrt{{6}^{2} + {2}^{2} + {3}^{2}}}$

$≡ \dfrac{6}{7}, \dfrac{2}{7}, \dfrac{3}{7}$

Now, given point $P\left(a, b, c\right)$ lies on the line $AB$ and at a distance of $21$ from point $A$

So, by using the equation of line $AB$ we get,

$⇒ \dfrac{a - 4}{\dfrac{6}{7}} = \dfrac{b + 6}{\dfrac{2}{7}} = \dfrac{c + 2}{\dfrac{3}{7}} = 21$

$⇒ \left(21 \times \dfrac{6}{7} + 4, \dfrac{2}{7} \times 21 - 6, \dfrac{3}{7} \times 21 - 2\right) ≡ \left(a, b, c\right)$

$⇒ \left(22, 0, 7\right) ≡ \left(a, b, c\right)$

Hence, the distance between $P\left(22, 0, 7\right) \text{ and } Q\left(4, - 12, 3\right)$ is given by, $\sqrt{324 + 144 + 16} = 22$