If the plane vec{r} = (lambda + mu)hat{i} + ( | vedclass

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(D) The given equation is $\vec{r} = \lambda(\hat{i} + 3\hat{k}) + \mu(\hat{i} + \hat{j} + 2\hat{k}) + 2\hat{j}$.Let $\vec{r} = x\hat{i} + y\hat{j} +

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$10/3$

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(D) The given equation is $\vec{r} = \lambda(\hat{i} + 3\hat{k}) + \mu(\hat{i} + \hat{j} + 2\hat{k}) + 2\hat{j}$.Let $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$. Then $x = \lambda + \mu$, $y = 2 + \mu$, $z = 3\lambda + 2\mu$.From $y = 2 + \mu$, we get $\mu = y - 2$.Substitute $\mu$ into $x = \lambda + \mu$: $x = \lambda + y - 2 \implies \lambda = x - y + 2$.Substitute $\lambda$ and $\mu$ into $z = 3\lambda + 2\mu$: $z = 3(x - y + 2) + 2(y - 2) = 3x - 3y + 6 + 2y - 4 = 3x - y + 2$.So, the equation of the plane is $3x - y - z = -2$, or $3x - y - z + 2 = 0$.For the intercept form, $3x - y - z = -2 \implies \frac{x}{-2/3} + \frac{y}{2} + \frac{z}{2} = 1$.Thus, $a = -2/3$, $b = 2$, $c = 2$.Then $a + b + c = -2/3 + 2 + 2 = -2/3 + 4 = 10/3$.

If the plane $\vec{r} = (\lambda + \mu)\hat{i} + (2 + \mu)\hat{j} + (3\lambda + 2\mu)\hat{k}$, where $\lambda$ and $\mu$ are parameters, intersects coordinate axes at points $(a, 0, 0), (0, b, 0), (0, 0, c)$, then $a + b + c = $

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