Let bar{a} = bar{i} + 2bar{j} + 3bar{k}, bar{ | vedclass

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(D) Let $\bar{r} = x\bar{i} + y\bar{j} + z\bar{k}$.Given the dot products:$x + 2y + 3z = 0$  $(1)$$2x - 3y + z = -2$  $(2)$$3x + y - 2z = 6$  $(3)$Sol

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$3$

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(D) Let $\bar{r} = x\bar{i} + y\bar{j} + z\bar{k}$.Given the dot products:$x + 2y + 3z = 0$  $(1)$$2x - 3y + z = -2$  $(2)$$3x + y - 2z = 6$  $(3)$Solving this system of linear equations:From $(1)$, $x = -2y - 3z$.Substitute into $(2)$: $2(-2y - 3z) - 3y + z = -2 \implies -7y - 5z = -2 \implies 7y + 5z = 2$  $(4)$Substitute into $(3)$: $3(-2y - 3z) + y - 2z = 6 \implies -5y - 11z = 6$  $(5)$Multiply $(4)$ by $5$ and $(5)$ by $7$: $35y + 25z = 10$ and $-35y - 77z = 42$.Adding these: $-52z = 52 \implies z = -1$.Substitute $z = -1$ into $(4)$: $7y - 5 = 2 \implies 7y = 7 \implies y = 1$.Substitute $y = 1, z = -1$ into $(1)$: $x + 2(1) + 3(-1) = 0 \implies x - 1 = 0 \implies x = 1$.Thus, $\bar{r} = \bar{i} + \bar{j} - \bar{k}$.Finally, calculate $\bar{r} \cdot (3\bar{i} + \bar{j} + \bar{k}) = (1)(3) + (1)(1) + (-1)(1) = 3 + 1 - 1 = 3$.

List-$I$	List-$II$
$A$. $a = 2\hat{i} + 3\hat{j} + 4\hat{k}, b = 3\hat{i} + 4\hat{j} + 2\hat{k}, c = 4\hat{i} + 2\hat{j} + 3\hat{k}$	$I$. $\triangle ABC$ is an equilateral triangle
$B$. $a = \hat{i} + 2\hat{j} + 3\hat{k}, b = 3\hat{i} + 4\hat{j} + 7\hat{k}, c = -3\hat{i} - 2\hat{j} - 5\hat{k}$	$II$. $\triangle ABC$ is an isosceles triangle
$C$. $a = 2\hat{i} - \hat{j} + \hat{k}, b = \hat{i} - 3\hat{j} - 5\hat{k}, c = -3\hat{i} - 4\hat{j} - 4\hat{k}$	$III$. $\triangle ABC$ is a right-angled triangle
$D$. $a = \hat{i} + \hat{j} + \hat{k}, b = \hat{i} + 2\hat{j} + 3\hat{k}, c = 2\hat{i} - \hat{j} + \hat{k}$	$IV$. $A, B, C$ are collinear

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