Let a line L passing through the point (1, 1, | vedclass

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(C) The direction vector $\vec{v}$ of line $L$ is given by the cross product of the two given vectors:$\vec{v} = (2\hat{i} + 2\hat{j} + \hat{k}) 	ime

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$100$

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(C) The direction vector $\vec{v}$ of line $L$ is given by the cross product of the two given vectors:$\vec{v} = (2\hat{i} + 2\hat{j} + \hat{k}) 	imes (\hat{i} + 2\hat{j} + 2\hat{k}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 2 & 1 \ 1 & 2 & 2 \end{vmatrix} = \hat{i}(4-2) - \hat{j}(4-1) + \hat{k}(4-2) = 2\hat{i} - 3\hat{j} + 2\hat{k}$.The equation of line $L$ passing through $(1, 1, 1)$ with direction vector $(2, -3, 2)$ is $\frac{x-1}{2} = \frac{y-1}{-3} = \frac{z-1}{2} = k$.Any point on line $L$ is $(2k+1, -3k+1, 2k+1)$.Let $P(a, b, c)$ be the foot of the perpendicular from the origin $(0, 0, 0)$ to line $L$. The vector $\vec{OP} = (2k+1, -3k+1, 2k+1)$ must be perpendicular to the direction vector of $L$, which is $(2, -3, 2)$.Thus, $2(2k+1) - 3(-3k+1) + 2(2k+1) = 0$.$4k + 2 + 9k - 3 + 4k + 2 = 0 \Rightarrow 17k + 1 = 0 \Rightarrow k = -1/17$.Substituting $k$ to find coordinates of $P$:$a = 2(-1/17) + 1 = 15/17$, $b = -3(-1/17) + 1 = 20/17$, $c = 2(-1/17) + 1 = 15/17$.Then $a + b + c = (15 + 20 + 15) / 17 = 50/17$.The value of $34(a + b + c) = 34 	imes (50/17) = 2 	imes 50 = 100$.

Let a line $L$ passing through the point $(1, 1, 1)$ be perpendicular to both the vectors $2\hat{i} + 2\hat{j} + \hat{k}$ and $\hat{i} + 2\hat{j} + 2\hat{k}$. If $P(a, b, c)$ is the foot of the perpendicular from the origin on the line $L$, then the value of $34(a + b + c)$ is:

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