If alpha and beta are scalars and vec{r} = (2 | vedclass

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(D) Given the vector equation of the plane: $\vec{r} = (2+\alpha-3\beta) \hat{i} + (\beta-3) \hat{j} + (2\alpha-5\beta-1) \hat{k}$.Let $\vec{r} = x \h

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$2x+y-z=2$

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(D) Given the vector equation of the plane: $\vec{r} = (2+\alpha-3\beta) \hat{i} + (\beta-3) \hat{j} + (2\alpha-5\beta-1) \hat{k}$.Let $\vec{r} = x \hat{i} + y \hat{j} + z \hat{k}$.Comparing the components, we have:$x = 2 + \alpha - 3\beta$  $(i)$$y = \beta - 3 \implies \beta = y + 3$  (ii)$z = 2\alpha - 5\beta - 1$  (iii)Substitute $\beta = y + 3$ into (iii):$z = 2\alpha - 5(y + 3) - 1$$z = 2\alpha - 5y - 15 - 1$$z = 2\alpha - 5y - 16$$2\alpha = z + 5y + 16 \implies \alpha = \frac{z + 5y + 16}{2}$.Now, substitute $\alpha$ and $\beta$ into $(i)$:$x = 2 + \left(\frac{z + 5y + 16}{2}\right) - 3(y + 3)$Multiply by $2$:$2x = 4 + z + 5y + 16 - 6y - 18$$2x = z - y + 2$$2x + y - z = 2$.

If $\alpha$ and $\beta$ are scalars and $\vec{r} = (2+\alpha-3\beta) \hat{i} + (\beta-3) \hat{j} + (2\alpha-5\beta-1) \hat{k}$ is the equation of a plane, then its equation in Cartesian form is:

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