Point $P(-6,3,0)$ to the line determined by the points $Q(3,-1,5)$ and $R(4,5,-2)$

A point $P(-6,3,0)$ to the line determined by the points $Q(3,-1,5)$ and $R(4,5,-2)$

The goal is to calculate the distance between the point and the line.

First find the line equation for the points $Q(3,-1,5)$and $R(4,5,-2)$

The direction vector $d=(4-3,5-(-1),-2-5)$

$d=(1,6,-7)$

The formula to find the line $\mathcal{L}$equation is as follows,

$r\left(t\right)=P_0+td$Where, $P_0$is the point and $d$is the direction vector.

Then, for $Q(3,-1,5)$and $d=(1,6,-7)$

$$\begin{gathered} r\left(t\right)=(3,-1,5)+t(1,6,-7) \\ r\left(t\right)=(3+t,-1+6t,5-7t) \end{gathered}$$

The formula for the distance is $\frac{||d\times P_{0}P||}{\Vert d\Vert }$

Let the point $P$is $P(-6,3,0)$

The point $P_0$ on the line equation $r\left(t\right)=(3+t,-1+6t,5-7t)$is $(3,-1,5)$

The direction vector $\overrightarrow{P_{0}P}=(-6-3,3-(-1),0-5)$

Then $\overrightarrow{P_{0}P}=(-9,4,-5)$

Take the direction vector $d=(1,6,-7)$

Substitute the values $d=(1,6,-7)$ and $\overrightarrow{P_{0}P}=(-9,4,-5)$ in the formula $\frac{||d\times \overrightarrow{P_{0}P}||}{\Vert d\Vert }$

Then the distance $=\frac{\Vert (1,6,-7)\times (-9,4,-5)\Vert }{\Vert (1,6,-7)\Vert }$

The cross product of $(1,6,-7)\times (-9,4,-5)$is calculated as follows,

$(1,6,-7)\times (-9,4,-5)=\left|\begin{array}{c}ijk\\ 16-7\\ -94-5\end{array}\right|$

$$\begin{gathered} =i(-30+28)-j(-5-63)+k(4+54) \\ =-2i+68j+58k \end{gathered}$$

Thus,

$$\begin{gathered} \text{Distance}=\frac{\Vert -2i+68j+58k\Vert }{\Vert (1,6,-7)\Vert } \\ \text{Distance}=\frac{\sqrt{4+4624+3364}}{\sqrt{1+36+49}} \\ \text{Distance}=\frac{\sqrt{7992}}{\sqrt{86}} \end{gathered}$$

On further simplification,

Distance $=\sqrt{\frac{3996}{43}}$

Distance $=9.64$ units

Thus, the distance between the point $P(-6,3,0)$ and the line joining $Q(3,-1,5)$ and $R(4,5,-2)$ is $9.64$ units.

Therefore, the answer is $9.64$ units.$R(4,5,-2)$