The distance of the point Q ( 0, 2, – 2 ) form the line passing...

Plotting the diagram of the given data we get,

Now, let $a, b, c$ be the DR's of the line perpendicular to \left(- 3\overset{^}{i} + 2\overset{^}{k}\right) + \lambda\left(2\overset{^}{i} + 3\overset{^}{j} + 5\overset{^}{k}\right)\text{and}\left(\overset{^}{i} - 2\overset{^}{j} + \overset{^}{k}\right) + \mu\left(- \overset{^}{i} + 3\overset{^}{j} + 2\overset{^}{k}\right)$.</p><p>$⇒ 2a + 3b + 5c = 0, - a + 3b + 2c = 0$</p><p>Now, solving$2a + 3b + 5c = 0 \text{ and } - a + 3b + 2c = 0$we get,</p><p>$⇒ \dfrac{a}{- 9} = \dfrac{b}{- 9} = \dfrac{c}{9}$</p><p>$⇒ \dfrac{a}{1} = \dfrac{b}{1} = \dfrac{c}{- 1}$</p><p>So, the equation will be,</p><p>$⇒ \dfrac{x - 5}{1} = \dfrac{y + 4}{1} = \dfrac{z - 3}{- 1} . . . \left(i\right)$</p><p>Let$R$be the foot of perpendicular from$Q\left(0, 2, - 2\right)\text{to}\left(i\right)$.</p><p>$⇒ \dfrac{x - 5}{1} = \dfrac{y + 4}{1} = \dfrac{z - 3}{- 1} = k$</p><p>$⇒ x = k + 5, y = k - 4, z = - k + 3$</p><p>$⇒ R ≡ \left(k + 5, k - 4, - k + 3\right)$</p><p>So, DR's of$QR$are \left(k + 5, k - 6, 5 - k\right)$.

Now, using perpendicular condition we get,

$⇒ 1\left(k + 5\right) + 1\left(k - 6\right) + \left(- 1\right)\left(5 - k\right) = 0$

$⇒ k = 2$

$⇒ R ≡ \left(7, - 2, 1\right)$

$⇒ QR = \sqrt{49 + 16 + 9}$

$⇒ QR = \sqrt{74}$