Let P be the plane passing through the points (5, 3, 0), (13, 3, -...

Given,

$P$ be the plane passing through the points $\left(5, 3, 0\right),$\left(13, 3, - 2\right)$and \left(1, 6, 2\right)$

Now, \overset{\to}{AB} = 8\overset{^}{i} - 2\overset{^}{k}, \overset{\to}{AC} = - 4\overset{^}{i} + 3\overset{^}{j} + 2\overset{^}{k} ,

So, normal vector to the plane will be,

\overset{\to}{AB} \times \overset{\to}{AC} = \left|\begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ 8 & 0 & - 2 \\ - 4 & 3 & 2\end{cases}\right|= 6\overset{^}{i} - 8\overset{^}{j} + 24\overset{^}{k}

Now equation of plane will be: $6x - 8y + 24z = d$

And given plane passes through \left(5, 3, 0\right)$</p><p>So,$6 \times 5 - 8 \times 3 + 24 \times 0 = d$$⇒ d = 6$

Hence, equation of plane will be,
$6x - 8y + 24z = 6 ⇒ 3x - 4y + 12z = 3$

Now using the formula distance of point $\left(3, 4, \alpha\right)$ from the plane $3x - 4y + 12z = 3$, we get,

$\dfrac{9 - 16 + 12\alpha - 3}{\sqrt{9 + 16 + 144}} = 2 ⇒ \alpha = 3$

Now again using the distance of point $\left(2, \alpha, a\right) ≡$\left(2, 3, a\right)$from the plane formula we get,</p><p>$\dfrac{3 \times 2 - 4 \times 3 + 12 \times a - 3}{13} = 3$</p><p>$⇒ 12a - 9 = 39$</p><p>$⇒ 12a = 48$</p><p>$⇒ a = 4$