Answer

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**Hint:**Here we can write the equation of the line as point is given $\left( {1, - 5,9} \right)$. And direction ratio is also given as $\left( {1,1,1} \right)$. So equation is written by

$\dfrac{{x - {x_1}}}{a} = \dfrac{{y - {y_1}}}{b} = \dfrac{{z - {z_1}}}{c}$ where $\left( {x,y,z} \right)$ are the points and $\left( {a,b,c} \right)$ are the direction ratios.Using this equation we find the coordinate points on the plane then finally we calculate distance between these two points which is our required answer.

**Complete step-by-step answer:**

Here, a plane is given $x - y + z = 5$ and a point anywhere is given $\left( {1, - 5,9} \right)$ which is to be measured along the line $x = y = z$. So if we draw the diagram, it seems like this:

So here, $ABCD$ is a plane satisfying $x - y + z = 5$ and $Q$ is the point on this plane and $P$ is the given point and $PQ$ is the line along $x = y = z$.

So we need to find $PQ$.

So for that we need to find the coordinates of $Q$.

Here the point $P\left( {{x_1},{y_1},{z_1}} \right) = \left( {1, - 5,9} \right)$ is given. And the point $Q\left( {x,y,z} \right)$ is assumed.

And we are given that $PQ$ is along the line $x = y = z$, that means we are given a direction ratio, that is $\left( {1,1,1} \right)$.

So equation of line $PQ$ must be written as

$\dfrac{{x - {x_1}}}{{{{\left( {dr} \right)}_1}}} = \dfrac{{y - {y_1}}}{{{{\left( {dr} \right)}_2}}} = \dfrac{{z - {z_1}}}{{{{\left( {dr} \right)}_3}}}$

Here $dr$ is the direction ratio.

So, $\dfrac{{x - 1}}{1} = \dfrac{{y + 5}}{1} = \dfrac{{z - 9}}{1} = \lambda $

Let this be equal to $\lambda $

So, here we can equate,

$

\dfrac{{x - 1}}{1} = \lambda \\

\dfrac{{y + 5}}{1} = \lambda \\

\dfrac{{z - 9}}{1} = \lambda \\

$

So,

$

x = \lambda + 1 \\

y = \lambda - 5 \\

z = \lambda + 9 \\

$

Here, $\left( {x,y,z} \right)$ lies on the plane $ABCD$ so it must satisfy the equation $x - y + z = 5$

So by putting $\left( {x,y,z} \right)$ values,

$

\lambda + 1 - \lambda + 5 + \lambda + 9 = 5 \\

\lambda = - 10 \\

$

So now we get $\lambda = - 10$.

Now,

$

x = \lambda + 1 = - 10 + 1 = - 9 \\

y = \lambda - 5 = - 10 - 5 = - 15 \\

z = \lambda + 9 = - 10 + 9 = - 1 \\

$

So, we got, $Q\left( {x,y,z} \right) = \left( { - 9, - 15, - 1} \right)$

Now by distance formula, we can find $PQ$

$PQ = \sqrt {{{\left( {x - {x_1}} \right)}^2} + {{\left( {y - {y_1}} \right)}^2} + {{\left( {z - {z_1}} \right)}^2}} $

Here, $P\left( {{x_1},{y_1},{z_1}} \right) = \left( {1, - 5,9} \right)$

And $Q\left( {x,y,z} \right) = \left( { - 9, - 15, - 1} \right)$

So,

$

PQ = \sqrt {{{\left( { - 9 - 1} \right)}^2} + {{\left( { - 15 + 5} \right)}^2} + {{\left( { - 1 - 9} \right)}^2}} \\

= \sqrt {{{\left( { - 10} \right)}^2} + {{\left( { - 10} \right)}^2} + {{\left( { - 10} \right)}^2}} \\

= \sqrt {100 + 100 + 100} \\

= \sqrt {300} \\

= 10\sqrt 3 \\

$

So, the distance of point $\left( {1, - 5,9} \right)$ from the plane $x - y + z = 5$ measured along the line $x = y = z$ is $10\sqrt 3 $.

**So, the correct answer is “Option B”.**

**Note:**If the equation of plane is given as $ax + by + cz + d = 0$ and only point $\left( {{x_1},{y_1},{z_1}} \right)$ is given, so the perpendicular distance of that point from plane $ax + by + cz + d = 0$ is given as

$\dfrac{{|a{x_1} + b{y_1} + c{z_1} + d|}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}$

This is valid when we need to find the perpendicular distance. If we need to find distance along any direction, then use the above method as stated.

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