A plane P contains the line of intersection of the plane → r · ( ^...

Given,

A plane $P$ contains the line of intersection of the plane \overset{\to}{r} \cdot \left(\overset{^}{i} + \overset{^}{j} + \overset{^}{k}\right) = 6 and \overset{\to}{r} \cdot \left(2\overset{^}{i} + 3\overset{^}{j} + 4\overset{^}{k}\right) = - 5,

Now let, ${P}_{1}: x + y + z - 6 = 0$

And ${P}_{2}: 2x + 3y + 4z + 5 = 0$

So, by family of planes concept we get,

$P: \left(x + y + z - 6\right) + \lambda\left(2x + 3y + 4z + 5\right) = 0$

Also given plane $P$ passes through $( 0, 2, - 2 )$

So, \left(0 + 2 - 2 - 6\right) + \lambda\left(4 + 6 - 8 + 5\right) = 0$</p><p>$⇒ - 6 + \lambda\left(3\right) = 0$</p><p>$⇒ \lambda = 2$</p><p>Hence, equation of plane will be,</p><p>$P: 5x + 7y + 9z + 4 = 0$</p><p>Now distance of point \left(12, 12, 18\right)$ from plane will be, \left(\dfrac{5\left(12\right) + 7\left(12\right) + 9\left(18\right) + 4}{\sqrt{25 + 49 + 81}}\right)$</p><p>And square of distance will be,</p><p>$= {\left(\dfrac{5\left(12\right) + 7\left(12\right) + 9\left(18\right) + 4}{\sqrt{25 + 49 + 81}}\right)}^{2}$</p><p>$= \dfrac{310 \times 310}{155} = 620