The point of intersection of the line joining | vedclass

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(D) The line passes through $A(1, 2, 1)$ and $B(2, -1, -1)$. The direction vector is $\vec{v} = B - A = (2-1)\bar{i} + (-1-2)\bar{j} + (-1-1)\bar{k} =

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$\frac{1}{7}(15\bar{i} - 10\bar{j} - 9\bar{k})$

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(D) The line passes through $A(1, 2, 1)$ and $B(2, -1, -1)$. The direction vector is $\vec{v} = B - A = (2-1)\bar{i} + (-1-2)\bar{j} + (-1-1)\bar{k} = \bar{i} - 3\bar{j} - 2\bar{k}$. The equation of the line is $\vec{r} = (1 + t)\bar{i} + (2 - 3t)\bar{j} + (1 - 2t)\bar{k}$.The plane passes through $P(1, 0, 0)$, $Q(0, 2, 0)$, and $R(0, 0, 3)$. The intercept form of the plane is $\frac{x}{1} + \frac{y}{2} + \frac{z}{3} = 1$, which simplifies to $6x + 3y + 2z = 6$.Substituting the line coordinates into the plane equation: $6(1 + t) + 3(2 - 3t) + 2(1 - 2t) = 6$.$6 + 6t + 6 - 9t + 2 - 4t = 6$.$14 - 7t = 6 \implies 7t = 8 \implies t = \frac{8}{7}$.Substituting $t = \frac{8}{7}$ into the line equation: $x = 1 + \frac{8}{7} = \frac{15}{7}$, $y = 2 - 3(\frac{8}{7}) = \frac{14 - 24}{7} = -\frac{10}{7}$, $z = 1 - 2(\frac{8}{7}) = \frac{7 - 16}{7} = -\frac{9}{7}$.The intersection point is $\frac{1}{7}(15\bar{i} - 10\bar{j} - 9\bar{k})$.

The point of intersection of the line joining the points $\bar{i} + 2\bar{j} + \bar{k}$ and $2\bar{i} - \bar{j} - \bar{k}$ and the plane passing through the points $\bar{i}, 2\bar{j}, 3\bar{k}$ is:

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