Let a line L_1 pass through the origin and be | vedclass

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(A) The direction vectors of $L_2$ and $L_3$ are $\vec{d}_2 = (1, 2, 2)$ and $\vec{d}_3 = (2, 2, 1)$.
The direction vector of $L_1$ is $\vec{d}_1 = \

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$4$

Vedclass · Answer

(A) The direction vectors of $L_2$ and $L_3$ are $\vec{d}_2 = (1, 2, 2)$ and $\vec{d}_3 = (2, 2, 1)$.
The direction vector of $L_1$ is $\vec{d}_1 = \vec{d}_2 	imes \vec{d}_3 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 2 \ 2 & 2 & 1 \end{vmatrix} = \hat{i}(2-4) - \hat{j}(1-4) + \hat{k}(2-4) = (-2, 3, -2)$.
Since $L_1$ passes through the origin,its equation is $\vec{r} = k(-2, 3, -2)$.
For the intersection of $L_1$ and $L_2$: $(-2k, 3k, -2k) = (3+t, 2t-1, 2t+4)$.
$-2k = 3+t$,$3k = 2t-1$,$-2k = 2t+4$.
From $3+t = 2t+4$,we get $t = -1$. Then $-2k = 3-1 = 2$,so $k = -1$.
The intersection point $P$ is $(-2(-1), 3(-1), -2(-1)) = (2, -3, 2)$.
Let the point on $L_3$ be $Q = (3+2s, 3+2s, 2+s)$.
The distance $PQ = \sqrt{17}$,so $PQ^2 = 17$.
$(3+2s-2)^2 + (3+2s+3)^2 + (2+s-2)^2 = 17$.
$(2s+1)^2 + (2s+6)^2 + s^2 = 17$.
$4s^2 + 4s + 1 + 4s^2 + 24s + 36 + s^2 = 17$.
$9s^2 + 28s + 20 = 0$.
$(9s+10)(s+2) = 0$,so $s = -2$ or $s = -10/9$.
Since $a \in Z$,we choose $s = -2$.
Then $Q = (3+2(-2), 3+2(-2), 2-2) = (-1, -1, 0)$.
Thus,$a = -1, b = -1, c = 0$.
$(a+b+c)^2 = (-1-1+0)^2 = (-2)^2 = 4$.

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