Let a line l pass through the origin and be p | vedclass

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(B) The direction vector of line $l$ is perpendicular to the direction vectors of $l_1$ and $l_2$. Let $\vec{v}_1 = \hat{i} + 2\hat{j} + 3\hat{k}$ and

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$5$

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(B) The direction vector of line $l$ is perpendicular to the direction vectors of $l_1$ and $l_2$. Let $\vec{v}_1 = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{v}_2 = 2\hat{i} + 2\hat{j} + \hat{k}$. The direction of $l$ is $\vec{v} = \vec{v}_1 	imes \vec{v}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 3 \ 2 & 2 & 1 \end{vmatrix} = \hat{i}(2-6) - \hat{j}(1-6) + \hat{k}(2-4) = -4\hat{i} + 5\hat{j} - 2\hat{k}$. Since $l$ passes through the origin,its equation is $\vec{r} = \gamma(-4\hat{i} + 5\hat{j} - 2\hat{k})$.For point $P$ (intersection of $l$ and $l_1$): $-4\gamma = 1 + \lambda$,$5\gamma = -11 + 2\lambda$,$-2\gamma = -7 + 3\lambda$. Solving these,we get $\gamma = -1$,so $P = (4, -5, 2)$.For point $Q$ on $l_2$,$Q = (-1 + 2\mu, 2\mu, 1 + \mu)$. The vector $\vec{PQ} = (-5 + 2\mu, 5 + 2\mu, -1 + \mu)$. Since $\vec{PQ} \perp \vec{v}_2$,we have $\vec{PQ} \cdot (2\hat{i} + 2\hat{j} + \hat{k}) = 0$,which gives $2(-5 + 2\mu) + 2(5 + 2\mu) + 1(-1 + \mu) = 0 \implies -10 + 4\mu + 10 + 4\mu - 1 + \mu = 0 \implies 9\mu = 1 \implies \mu = 1/9$.Thus,$Q = (-1 + 2/9, 2/9, 1 + 1/9) = (-7/9, 2/9, 10/9)$.Then $9(\alpha + \beta + \gamma) = 9(-7/9 + 2/9 + 10/9) = 9(5/9) = 5$.

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