Question

find the point of intersection p of the line \\( \mathbf{r}(t) = \langle 1, -3, 2 \
angle + t \langle -3, 2, -2 \
angle \\) with the plane \\( 3z - (4x + y) = 5 \\): \\( p = \left( \square, \square, \square \
ight) \\)

Explanation:

Step1: Extract line's parametric equations

The line is \(\mathbf{r}(t)=\langle 1, - 3,2
angle+t\langle - 3,2,-2
angle\). So the parametric equations are:
\(x = 1-3t\), \(y=-3 + 2t\), \(z = 2-2t\)

Step2: Substitute into plane equation

The plane equation is \(3z-(4x + y)=5\). Substitute \(x,y,z\) from step 1:
\[

$$\begin{align*} 3(2 - 2t)-(4(1-3t)+(-3 + 2t))&=5\\ 6-6t-(4-12t - 3+2t)&=5\\ 6-6t-(1 - 10t)&=5\\ 6-6t - 1+10t&=5\\ 5 + 4t&=5\\ 4t&=0\\ t&=0 \end{align*}$$

\]

Step3: Find intersection point

Substitute \(t = 0\) into parametric equations:
\(x=1-3(0)=1\), \(y=-3 + 2(0)=-3\), \(z=2-2(0)=2\)

Answer:

\(P=(1, - 3,2)\)