As in Problem 24, find the equations of the line intersections of the planes in Problem 23. Find the distance from the point (1,0,0) to the line.

As per Problem 23, the planes are$2x+y-2z=3$ and $3x-6y-2z=4$.

The given point is $\left(1,0,0\right)$

In order to find a point that lies on both the planes, put $z=0$.

Then equations of the planes become$2x+y=3$ and $3x-6y=4$.

Solving the two equations using elimination method, it implies

$x=\frac{22}{15}$and $y=\frac{1}{15}$

Thus, the common point is $\left(\frac{22}{15},\frac{1}{15},0\right)$

For the given planes, normal vectors are:$2\hat{i}+\hat{j}-2\hat{k}$

and$3\hat{i}-6\hat{j}-2\hat{k}$ respectively.

The cross product of the two normal vectors:

$$\begin{gathered} N=\left[\begin{array}{ccc}i& j& k\\ 2& 1& -2\\ 3& -6& -2\end{array}\right] \\ =\left(-2-12\right)\hat{i}-\left(-4+6\right)\hat{j}+\left(-12-3\right)\hat{k} \\ =-14\hat{i}-2\hat{j}-15\hat{k} \end{gathered}$$

Now, the equation of line of intersection of the planes through$\left(\frac{22}{15},\frac{1}{15},0\right)$ and parallel to$-14\hat{i}-2\hat{j}-15\hat{k}$ is

$\frac{x-\frac{22}{15}}{-14}=\frac{y-\frac{1}{15}}{-2}=\frac{z-0}{-15}$

So the equations are

$\frac{x-\frac{22}{15}}{-14}=\frac{y-\frac{1}{15}}{-2}=\frac{z}{-15}$

And parameteric form of equations is

$r=\left(\frac{22}{15}\right)\hat{i}+\left(\frac{1}{15}\right)\hat{j}+\left(-14\hat{i}-2\hat{j}-15\hat{k}\right)t$

The point is$P=\left(1,0,0\right)$

The line is$r=\left(\frac{22}{15}\right)\hat{i}+\left(\frac{1}{15}\right)\hat{j}+\left(-14\hat{i}-2\hat{j}-15\hat{k}\right)t$

Here,

$$\begin{gathered} Q=\frac{22}{15}\hat{i}+\frac{1}{15}\hat{j} \\ =\left(\frac{22}{15},\frac{1}{15},0\right) \end{gathered}$$

And $A=-14\hat{i}-2\hat{j}-15\hat{k}$

So

$$\begin{gathered} \left|A\right|=\sqrt{{\left(-14\right)}^2+{\left(-2\right)}^2+{\left(-15\right)}^2} \\ =\sqrt{425} \end{gathered}$$

Unit vector along the line is $u=\frac{-14\hat{i}-2\hat{j}-15\hat{k}}{\sqrt{425}}$

Also,

$$\begin{gathered} \overrightarrow{PQ}=\left(\frac{22}{15},\frac{1}{15},0\right)-\left(1,0,0\right) \\ =\left(\frac{7}{15},\frac{1}{15},0\right) \\ =\frac{7}{15}\hat{i}\frac{1}{15}\hat{j} \end{gathered}$$

The distance of the point$P=\left(1,0,0\right)$ to the line is

$$\begin{gathered} \left|\overrightarrow{PQ}\times u\right|=\left|\left(\frac{7}{15}\hat{i}+\frac{1}{15}\hat{j}\right)\times \frac{-14\hat{i}-2\hat{j}-15\hat{k}}{\sqrt{425}}\right| \\ =\frac{1}{\sqrt{425}}\left|\left(\frac{7}{15}\hat{i}+\frac{1}{15}\hat{j}\right)\times \left(-14\hat{i}-2\hat{j}-15\hat{k}\right)\right| \\ =\frac{1}{\sqrt{425}}\left|-\hat{i}+7\hat{j}\right| \\ =\frac{1}{\sqrt{425}}\left(\sqrt{{\left(-1\right)}^2+7^2}\right) \end{gathered}$$

So,

$$\begin{gathered} \left|\overrightarrow{PQ}\times u\right|=\frac{\sqrt{50}}{\sqrt{425}} \\ =\sqrt{\frac{2}{17}} \end{gathered}$$

Hence, the distance from the point (1,0,0) to the line is$\sqrt{\frac{2}{17}}$ units.