If the lines (x - 1)/(1) = (y - 2)/(2) = (z + 3)/(1) and (x -...

Let

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

Point on ${L}_{1} ≡ \left(\lambda + 1, 2\lambda + 2, \lambda - 3\right)</p><p>And,</p><p>${L}_{2}: $\dfrac{x - a}{2} = \dfrac{y + 2}{3} = \dfrac{z - 3}{1} =$\mu$</p><p>Point on${L}_{2} ≡ \left(2\mu + a, 3\mu - 2, \mu + 3\right)$</p><p>For finding$P$, we must have</p><p>$\lambda - 3 = \mu + 3 ⇒ \lambda = \mu + 6 . . . \left(i\right)

$2\lambda + 2 = 3\mu - 2 ⇒ 2\lambda = 3\mu - 4 . . . \left(ii\right)$

Solving \left(i\right) \text{and }\left(ii\right), we get</p><p>\lambda = 22$and$\mu = 16$</p><p>Therefore,</p><p>$P ≡ \left(23, 46, 19\right)$</p><p>And,</p><p>$2\mu + a = \lambda + 1$</p><p>$⇒ 32 + a = 22 + 1$</p><p>$⇒ a = - 9$</p><p>Distance of point$P\left(23, 46, 19\right) from the plane $z + 9 = 0$ is

$= \left|\dfrac{19 + 9}{\sqrt{{0}^{2} + {0}^{2} + {1}^{2}}}\right|$ = 28$units</p>$