Explanation

(d)
Two given lines intersect, if
$7\widehat{\dot{\mathbf{i}}} + 10\widehat{\dot{\mathbf{j}}} + 13\widehat{\mathbf{k}} + s\left( 2\widehat{\dot{\mathbf{i}}} + 3\widehat{\dot{\mathbf{j}}} + 4\widehat{\mathbf{k}} \right)$
$= 3\widehat{\dot{\mathbf{i}}} + 5\widehat{\dot{\mathbf{j}}} + 7\widehat{\mathbf{k}} + t(\widehat{\dot{\mathbf{i}}} + 2\widehat{\dot{\mathbf{j}}} + 3\widehat{\mathbf{k}})$
$\Longrightarrow (7 + 2s)\widehat{\dot{\mathbf{i}}} + (10 + 3s)\widehat{\dot{\mathbf{j}}} + (13 + 4s)\widehat{\mathbf{k}}$
$= (3 + t)\widehat{\dot{\mathbf{i}}} + (5 + 2t)\widehat{\dot{\mathbf{j}}} + (7 + 3t)\widehat{\mathbf{k}}$
$\Longrightarrow 7 + 2s = 3 + t$
$\Longrightarrow 2s - t = - 4$ …(i)
$10 + 3s = 5 + 2t$
$\Longrightarrow 3s - 2t = - 5$ …(ii)
$and\ 13 + 4s = 7 + 3t$
$\Longrightarrow 4s - 3t = - 6$ …(iii)
On solving Eqs. (i) and (iii), we get
$s = - 3,\ t = - 2$
$\therefore\$Required line is
$7\widehat{\dot{\mathbf{i}}} + 10\widehat{\dot{\mathbf{j}}} + 13\widehat{\mathbf{k}} + ( - 3)\lbrack 2\widehat{\dot{\mathbf{i}}} + 3\widehat{\dot{\mathbf{j}}} + 4\widehat{\mathbf{k}}\rbrack$
$\Longrightarrow \ \widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} + \widehat{\mathbf{k}}$ is the required line.