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## Homework Statement

Find the value of k such that no solutions are shared between the planes

x + 2y + kz = 6

and

3x + 6y + 8z = 4

## Homework Equations

## The Attempt at a Solution

I figured that if I calculated [itex]\vec{n_1} \times \vec{n_2} = \vec{0}[/itex], and then take the magnitude of this vector equation, I could procure the correct k-value.

Here are some parts of my work:

[itex]\vec{n_1} \times \vec{n_2} = (16-6k)\hat{i} + (3k-8)\hat{k}[/itex]

[itex]0=320 -240k + 39k^2[/itex]

[itex]k = \frac{120 \pm 8 \sqrt{30}}{39}[/itex]

When I graphed the two planes, it was quite conspicuous that the two planes were not parallel.

What is wrong with my work and reasoning?