Let P _ 1 be the plane 3x - y - 7z = 11 and P _ 2 be the plane...

Given,

${P}_{1}$ be the plane $3x - y - 7z = 11$,

${P}_{2}$ be the plane passing through the points $( 2, - 1, 0 )$, $( 2, 0, - 1 )$, and $( 5, 1, 1 )$,

Now finding the equation of plane ${P}_{2}$, we get

\left|$\begin{cases}x - 2 & y + 1 & z \\ 2 - 2 & 0 - \left(- 1\right) & - 1 - 0 \\ 5 - 2 & 1 - 0 & 1 - \left(- 1\right)\end{cases}$\right| = 0$</p><p>$⇒ \left|$\begin{cases}x - 2 & y + 1 & z \\ 0 & 1 & - 1 \\ 3 & 1 & 2\end{cases}$\right| = 0$</p><p>$⇒ 3\left(x - 2\right) + 3\left(- y - 1\right) - 3z = 0$</p><p>$⇒ x - 2 - y - 1 - z = 0$</p><p>$⇒ x - y - z = 3$</p><p>Now direction ratio of line of intersection of planes$3x - y - 7z = 11 \text{ and } x - y - z = 3is given by,<br />\left|\begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ 3 & - 1 & - 7 \\ 1 & - 1 & - 1\end{cases}\right|
= \overset{^}{i} ( 1 - 7 ) - \overset{^}{j} ( - 3 + 7 ) + \overset{^}{k} ( - 3 + 1 )
= - 6\overset{^}{i} - 4\overset{^}{j} - 2\overset{^}{k}

Or direction ratio is given by 3\overset{^}{i} + 2\overset{^}{j} + \overset{^}{k}

Now finding the point on the line of intersection by taking $z = 0$ in both plane,

We get, $3x - y = 11 \text{ and } x - y = 3$

Now solving above two lines we get,

$( 4, 1, 0 )$ is a point on line of intersection

Hence, the equation of line will be,

$\dfrac{x - 4}{3} = \dfrac{y - 1}{2} = \dfrac{z}{1} = \lambda$

Now any point on the line is given by,

\left(3\lambda + 4, 2\lambda + 1, \lambda\right)$</p><p>Now finding the distance of the point \left(7, 4, - 1\right)$ from the line,

Now from above diagram using perpendicular condition we get,

$3 ( 3\lambda - 3 ) + 2 ( 2\lambda - 3 ) + 1 ( \lambda + 1 ) = 0$

$⇒ 14\lambda = 14 ⇒ \lambda = 1$

Hence, the point is \left(7, 3, 1\right)$and comparing with \left(\alpha, \beta, \gamma\right)$ we get, $\alpha + \beta + \gamma = 11$