The distance of the point P(4, 6, - 2) from the line passing...

The equation of line passing through the point $\left(- 3, 2, 3\right)$ and parallel to a line with direction ratios $3, 3, - 1$ will be,

$\dfrac{x + 3}{3} = \dfrac{y - 2}{3} = \dfrac{z - 3}{- 1} = \lambda$

Now let any point on the line will be, $M\left(3\lambda - 3, 3\lambda + 2, 3 - \lambda\right)</p><p>We know the direction ratios of the line joining the points \left({x}_{1}, {y}_{1}, {z}_{1}\right) and \left({x}_{2}, {y}_{2}, {z}_{2}\right) is given by</p><p>\left({x}_{2} - {x}_{1}, {y}_{2} - {y}_{1}, {z}_{2} - {z}_{1}\right)</p><p>Therefore, the direction ratios of line PM,</p><p>D.R of$PM \left(3\lambda - 7, 3\lambda - 4, 5 - \lambda\right)$</p><p>Since,$PM$is perpendicular to the line</p><p>$⇒ 3\left(3\lambda - 7\right) + 3\left(3\lambda - 4\right) - 1\left(5 - \lambda\right) = 0$</p><p>$⇒ \lambda = 2$</p><p>$⇒ M\left(3, 8, 1\right)

Now distance$of point P from point M,</p><p>⇒ PM = \sqrt{{\left(3 - 4\right)}^{2} + {\left(8 - 6\right)}^{2} + {\left(1 + 2\right)}^{2}}$$</p><p>$⇒ PM = \sqrt{14}$</p>$