Answer

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**Hint:**To solve this question, we will use the concept of finding the equation of a line parallel to a given line and passing through a given point and also, we will use the distance formula.

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

Given that,

Equation of plane = $\left( {\hat i - \hat j + \hat k} \right) = 5$

Equation of line = $r = \hat i + \hat j + \hat k$

Given passing point = (1, -5, 9).

First, we will convert these vectors form into Cartesian forms,

So,

Equation of plane in cartesian form will be,

$ \Rightarrow x - y + z = 5$ ……. (i)

Equation of line in cartesian form will be,

$ \Rightarrow x = y = z$ ……… (ii)

According to the question,

The line through which the point (1, -5, 9) is passing, is parallel to the line x = y = z.

As we know that, the equation of a line parallel to a given line and passing through a given point is given by,

$ \Rightarrow \dfrac{{x - {x_1}}}{a} = \dfrac{{y - {y_1}}}{b} = \dfrac{{z - {z_1}}}{c}$,

Where a, b and c are the direction cosines of the given line and $\left( {{x_1},{y_1},{z_1}} \right)$ is the given point.

So, the equation of the line parallel to the line $x = y = z$ and passing through the point (1, -5, 9) will be,

$ \Rightarrow \dfrac{{x - 1}}{1} = \dfrac{{y + 5}}{1} = \dfrac{{z - 9}}{1} = \lambda $ [$\lambda $ be any scalar value]

Solving this, we will get

$ \Rightarrow x = \lambda + 1,y = \lambda - 5,z = \lambda + 9$

As we know that, this line is intersecting the given plane, which means this equation must satisfy the equation of plane.

Put these values of x, y and z in equation (i),

$ \Rightarrow \lambda + 1 - \left( {\lambda - 5} \right) + \lambda + 9 = 5$

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

$ \Rightarrow \lambda + 10 = 0$

$ \Rightarrow \lambda = - 10$

Now, putting this value in the values of x, y and z, we will get

$ \Rightarrow x = - 10 + 1 = - 9,$

$ \Rightarrow y = - 10 - 5 = - 15,$

$ \Rightarrow z = - 10 + 9 = - 1$

Thus, the point on the plane is (-9, -15, -1).

Now, we will find the distance between the points (1, -5, 9) and (-9, -15, -1) by using the distance formula,

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

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

$ \Rightarrow d = \sqrt {100 + 100 + 100} $

$ \Rightarrow d = \sqrt {300} $

$ \Rightarrow d = 10\sqrt 3 $ units.

Hence, the distance of the point (1, -5, 9), from the planar $\left( {\hat i - \hat j + \hat k} \right) = 5$ measured along the line $r = \hat i + \hat j + \hat k$ is $10\sqrt 3 $ units.

**Note:**Whenever we ask such type of question, we also have to remember that the distance of the point $\left( {{x_1},{y_1},{z_1}} \right)$ from the plane $ax + by + cz + d = 0$ is given by $\dfrac{{\left| {a{x_1} + b{y_1} + c{z_1} + d} \right|}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}$ and this is also called the foot of the perpendicular from $\left( {{x_1},{y_1},{z_1}} \right)$ to the plane $ax + by + cz + d = 0$

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