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# Find an equation of the plane.The plane that passes through the point $(6, -1, 3)$ and contains the line with symmetric equations $\frac{x}{3} = y + 4 = \frac{z}{2}$

## $x-y-z=4$

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Yeah. In the question they're asking to find the equation of the plane that passes through the 0.6 minus 13 and contains the line with symmetric equations. X by three were two, Y plus 40 quarters it by two, rearranging this equation and assigning this tool constant value, we can write it as X by three Is equal to Y-plus four is equal to zero x 2 is equal to T see it is equal to x equal to treaty, Then by equal to T -4 And that equal to two T. And the constant attached to these coordinates are zero minus four and zero. And the coefficients For these coordinates at three, one and 2 respectively. So a point in the plane is given, say it is called B Its value is 6 -1, 3. And another point in the plane considering no coordinates given in the question is 0 -4 and zero. That is the constant values for each of the coordinates and the normal victor passes to the point 312 in the plane. That is a coefficient of the coordinates. Now, in order to find the equation of the plane to find a line passing through the plane that they speak you which has the direction vector equal to 0 -6, icap plus minus four plus 1. Jacob Less 0 -3, Caykur, it is equal to -6. I got -3 Jacob -3. Jacob. Therefore the cross productive PQ and n victor is Ichabod Jacob, Jacob and the value of PQ is -6 -3 -3 and the value of In is 312. Mhm. Therefore the value of the cross product is equal to yeah Mhm minus three. Icap this tree Jacob Plus three K Cab. Therefore the equation of the plane is mhm. Yeah minus three X minus six. Last three Y plus one Last three. Zed -3. equal to zero considering point. Okay. Mhm. 6 -1. 3. Yeah that is given in the question. This is the point B. So this is equal to after solving the equation we get x minus y minus Z. Is equal to fool and this is the required answer. Are the given question? Yeah and they put this is the equation of the plane.

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