Answer

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**Hint:**

Here we will first find the value of the vector from the given point to the given line. Then we will find the directional vector of the line and perform its dot product to get the value of \[t\]. Then we will put the value of \[t\] in the obtained vector and find its magnitude which is equal to the distance between the given point and the line.

**Complete step by step solution:**

The given point is \[A\left( {3, - 5,5} \right)\] and the line L is \[L:x = 2 + 3t,y = 1 - 2t,z = - 1 + t\].

Let a point P is on the given line L. Therefore the coordinates of the point is \[P\left( {2 + 3t,1 - 2t, - 1 + t} \right)\].

So, firstly we will find the value of the vector \[\mathop {AP}\limits^ \to \] which is the vector from the point to the line. Therefore, we get

\[\mathop {AP}\limits^ \to = \left( {2 + 3t - 3,1 - 2t + 5, - 1 + t - 5} \right)\]

Adding and subtracting the like terms, we get

\[\Rightarrow \mathop {AP}\limits^ \to = \left( {3t - 1,6 - 2t,t - 6} \right)\]

Now we will find the directional vector of the given line using the equation of the given line L. therefore, the directional vector is equal to

\[\vec l = \left( {3, - 2,1} \right)\]

We know that the directional vector of the line will be perpendicular to the vector \[\mathop {AP}\limits^ \to \]. Therefore, the dot product of the vector will be equal to zero. Therefore, we get

\[ \Rightarrow \mathop {AP}\limits^ \to \cdot \vec l = 0\]

\[ \Rightarrow \mathop {AP}\limits^ \to \cdot \vec l = \left( {3t - 1,6 - 2t,t - 6} \right) \cdot \left( {3, - 2,1} \right) = 0\]

\[ \Rightarrow \mathop {AP}\limits^ \to \cdot \vec l = 3\left( {3t - 1} \right) - 2\left( {6 - 2t} \right) + 1\left( {t - 6} \right) = 0\]

Now by solving this equation we will get the value of \[t\]. Therefore, we get

\[ \Rightarrow 9t - 3 - 12 + 4t + t - 6 = 0\]

\[ \Rightarrow 14t - 21 = 0\]

\[ \Rightarrow t = \dfrac{{21}}{{14}} = \dfrac{3}{2}\]

Now we will put the value of \[t\] in the vector \[\mathop {AP}\limits^ \to \] and find its magnitude to get the shore=test distance between the point and the given line L. therefore, we get

\[\mathop {AP}\limits^ \to = \left( {3 \times \dfrac{3}{2} - 1,6 - 2 \times \dfrac{3}{2},\dfrac{3}{2} - 6} \right)\]

\[ \Rightarrow \mathop {AP}\limits^ \to = \left( {\dfrac{7}{2},3, - \dfrac{9}{2}} \right)\]

Now we will find the magnitude of the above vector. Therefore, we get

\[\left\| {\mathop {AP}\limits^ \to } \right\| = \sqrt {{{\left( {\dfrac{7}{2}} \right)}^2} + {3^2} + {{\left( { - \dfrac{9}{2}} \right)}^2}} \]

Simplifying the expression, we get

\[\Rightarrow \left\| {\mathop {AP}\limits^ \to } \right\| = \sqrt {\dfrac{{49}}{4} + 9 + \dfrac{{81}}{4}} = \sqrt {\dfrac{{49 + 36 + 81}}{4}} \]

Adding the terms, we get

\[ \Rightarrow \left\| {\mathop {AP}\limits^ \to } \right\| = \sqrt {\dfrac{{166}}{4}} = \sqrt {\dfrac{{83}}{2}} \]

**Hence the distance from the point \[A\left( {3, - 5,5} \right)\] to the line \[x = 2 + 3t,y = 1 - 2t,z = - 1 + t\] is equal to \[\sqrt {\dfrac{{83}}{2}} \] units.**

**Note:**

Vector is the geometric object that has both the magnitude and the direction of an object. So while calculating the equation of a line vector we should know that it is equal to the difference between the final point vector and the starting point vector of that line. Here we have to keep in mind that while expanding the dot product of vectors is related to cos function not the sin function. Sin function is related to the cross product of the vectors. Also we have to remember that the dot product of the perpendicular vectors is always zero as the angle between them is \[{90^0}\] and the dot product is related to cos function. However, the cross product of the parallel vectors is always zero as the angle between the parallel vectors is \[{0^0}{\rm{or18}}{{\rm{0}}^0}\] and cross product is related to the sin function.

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