A line L passes through the points hat{i}+2 h | vedclass

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(B) The equation of the line $L$ passing through $(1, 2, 1)$ and $(-2, 0, 3)$ is given by $\frac{x-1}{-3} = \frac{y-2}{-2} = \frac{z-1}{2} = \lambda$.

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$-8 \hat{i}-4 \hat{j}+7 \hat{k}$

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(B) The equation of the line $L$ passing through $(1, 2, 1)$ and $(-2, 0, 3)$ is given by $\frac{x-1}{-3} = \frac{y-2}{-2} = \frac{z-1}{2} = \lambda$. Any point on this line is $A(-3\lambda + 1, -2\lambda + 2, 2\lambda + 1)$. The plane $P$ passes through $(0, 0, 0)$, $(0, 0, 4)$, and $(2, 1, 0)$. The equation of the plane is given by the determinant $\begin{vmatrix} x & y & z \ 0 & 0 & 4 \ 2 & 1 & 0 \end{vmatrix} = 0$. Expanding the determinant, we get $-4(x - 2y) = 0$, which simplifies to $x - 2y = 0$. Since point $A$ lies on the plane, substitute the coordinates of $A$ into the plane equation: $(-3\lambda + 1) - 2(-2\lambda + 2) = 0$. $-3\lambda + 1 + 4\lambda - 4 = 0 \Rightarrow \lambda - 3 = 0 \Rightarrow \lambda = 3$. Substituting $\lambda = 3$ into the coordinates of $A$: $x = -3(3) + 1 = -8$, $y = -2(3) + 2 = -4$, $z = 2(3) + 1 = 7$. Thus, the point is $-8\hat{i} - 4\hat{j} + 7\hat{k}$.

$A$ line $L$ passes through the points $\hat{i}+2 \hat{j}+\hat{k}$ and $-2 \hat{i}+3 \hat{k}$. $A$ plane $P$ passes through the origin and the points $4 \hat{k}, 2 \hat{i}+\hat{j}$. The point where the line $L$ meets the plane $P$ is

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