Let P_{1}: vec{r} cdot(2 hat{i} + hat{j} - 3 | vedclass

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(B) The equation of plane $P_{1}$ is $2x + y - 3z = 4$. The normal vector is $\vec{n}_{1} = 2\hat{i} + \hat{j} - 3\hat{k}$.The plane $P_{2}$ passes th

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$28$

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(B) The equation of plane $P_{1}$ is $2x + y - 3z = 4$. The normal vector is $\vec{n}_{1} = 2\hat{i} + \hat{j} - 3\hat{k}$.The plane $P_{2}$ passes through points $A(2, -3, 2)$,$B(2, -2, -3)$,and $C(1, -4, 2)$.Vectors in the plane are $\vec{AB} = (2-2)\hat{i} + (-2+3)\hat{j} + (-3-2)\hat{k} = \hat{j} - 5\hat{k}$ and $\vec{AC} = (1-2)\hat{i} + (-4+3)\hat{j} + (2-2)\hat{k} = -\hat{i} - \hat{j}$.The normal vector $\vec{n}_{2}$ to $P_{2}$ is $\vec{AB} 	imes \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 0 & 1 & -5 \ -1 & -1 & 0 \end{vmatrix} = \hat{i}(0 - 5) - \hat{j}(0 - 5) + \hat{k}(0 + 1) = -5\hat{i} + 5\hat{j} + \hat{k}$.The direction ratios of the line of intersection are given by $\vec{n}_{1} 	imes \vec{n}_{2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 1 & -3 \ -5 & 5 & 1 \end{vmatrix} = \hat{i}(1 + 15) - \hat{j}(2 - 15) + \hat{k}(10 + 5) = 16\hat{i} + 13\hat{j} + 15\hat{k}$.Comparing with $16, \alpha, \beta$,we get $\alpha = 13$ and $\beta = 15$.Therefore,$\alpha + \beta = 13 + 15 = 28$.

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